Approximation Algorithms for Geometric Intersection Graphsssen/geomschool/nandy/GEO...Introduction Rectangle Intersection GraphsUnit Disk GraphDiscrete Piercing Set for Unit Disks

Introduction Rectangle Intersection Graphs Unit Disk Graph Discrete Piercing Set for Unit Disks

Approximation Algorithms for GeometricIntersection Graphs

Subhas C. Nandy([email protected])

Advanced Computing and Microelectronics UnitIndian Statistical Institute

Kolkata 700108, India.


Outline

1 Introduction

2 Rectangle Intersection GraphsMaximum Independent SetMinimum Clique Cover

3 Unit Disk Graph

4 Discrete Piercing Set for Unit Disks


Introduction

Geometric Intersection Graph

Consider a set of objects distributed in IRd .

The geometric intersection graph G (V ,E ) with these set ofobjects is an undirected graph where

• Each node in V corresponds to a distinct object in this set.

• Between a pair of nodes vi , vj ∈ V there is an edge eij (∈ E )if the corresponding objects intersect.

Example: Interval graph, circular arc graph, rectangle intersectiongraph, unit disk graph etc.


Introduction

Usefulness:

Many practical problems are formulated as an optimizationproblem of some geometric intersection graph.

In VLSI Physical Design rectangle intersection graphs play amajor role.

In wireless communication disk graphs play important role.

In cartography, intersection graphs of rectangles/circles playimportant role.


Problems of Interest

Characterization Problem Given a graph G , test whether it is theintersection of a set of objects of some desired shape.

Solving some useful optimization problems

We discuss on the existing results on the characterization problemsof different intersection graphs, and available algorithms for solvingthe optimization problems.

More specifically, we concentrate on the approximation algorithmsfor some important optimization problems on geometricintersection graphs.


Existing results on Characterization Problems

Testing whether a given graph G = (V ,E ) is an IntervalGraph needs O(|V |+ |E |) timeM. C. Golumbic, Algorithmic Graph Theory and Perfect Graphs, 1979

Testing whether a given graph G is a Rectangle IntersectionGraph is NP-complete.M. B. Cozzens, Higher and Multidimensional Analogues of Interval Graphs, 1981.

Testing whether a given graph G is a Unit Disk Graph isNP-complete.H. Breu, D. G. Kirkpatrick, Unit disk graph recognition is NP-hard, 1998.


Existing Algorithmic results on Interval Graphs andCircular Arc Graphs

Interval Graph

A

BC

D

E FG

A

B

C

D

E

F

G

Circular-Arc Graph

AB

C

D

E

F

GH

A

B

C

D E

F

G

H

Results:

Both these graphs can be recognized in O(n) time

Largest clique and Minimum clique cover can be computedin O(n log n) time.

Maximum independent set can be computed in O(n log n)time.

There exists an output sensitive algorithm (time complexityO(n log k)) for the maximum independent set of intervalgraph 1

1J. Snoeyink, Maximum independent set of intervals by divide prune andconquer, CCCG 2007


Definition

Any graph can be represented as the intersection graph of a set ofrectangles in d dimensional plane where d ≥ 1.The boxicity of a graph G is the minimum value of d .

Boxicity-2 Graphs are called the Rectangle Intersection Graphs

A

B

C

D

E

F

A

B

C

D

EF

Assumption: The rectangle representation of the graph is given.


Important Results on Boxicity of a Graph

1. A graph has boxicity one if and only if it is an interval graph.

2. Every outerplanar graph has boxicity at most two.

3. Every planar graph has boxicity at most three.

4. If a bipartite graph has boxicity two, it can be represented asan intersection graph of axis-parallel line segments in theplane.

3. The upper bound of the boxicity d is d ≤ τ + 1 where τ is thetree-width τ of the graph.


Largest Clique of Rectangle Intersection Graph

Let G = (V ,E ) be a rectangle intersection graph, and thecorresponding rectangle representation is given.

Algorithm: (* Use Plane Sweep *)

Process the horizontal boundariesof the rectangles in top-to-bottomorder.

An interval tree is used to maintainthe sweep line status. Its eachelement represents a clique so fardetected.

With each clique, the count ofrectangle lying on it is stored.

12 30 23 21 0



Let G = (V ,E ) be a rectangle intersection graph, and thecorresponding rectangle representation is given.

Algorithm: (* Use Plane Sweep *)

Process the horizontal boundariesof the rectangles in top-to-bottomorder.

An interval tree is used to maintainthe sweep line status. Its eachelement represents a clique so fardetected.

With each clique, the count ofrectangle lying on it is stored.

12 30 23 21 0


When a top boundary of a rectangle isprocessed,

Two clique-intervals get split, and thecount of all the cliques inside the rangeof the rectangle is increased by one.

When a bottom boundary is processed,

Report the largest clique contained inthis rectangle.

At most two clique-intervals getmerged with its neighboring interval.

The count of all the cliques inside therange of the rectangle is decreased byone.


When a top boundary of a rectangle isprocessed,

Two clique-intervals get split, and thecount of all the cliques inside the rangeof the rectangle is increased by one.

When a bottom boundary is processed,

Report the largest clique contained inthis rectangle.

At most two clique-intervals getmerged with its neighboring interval.

The count of all the cliques inside therange of the rectangle is decreased byone.



Result: The number of maximal cliques in the graph can be O(n2)in the worst case.

Result: The largest clique can be computed in O(n log n) time.

Application: Given a set of points in 2D and a rectangle ofspecified size, report the position of the rectangle on the planesuch that it can contain the maximum/minimum number of points


Hardness results on the optimization problems forRectangle Intersection Graph

Hardness result: Both minimum clique cover and maximumindependent set problems for a rectangle intersectiongraph are NP-hard 2.

Approximation Result: Both minimum clique cover3 and maximumindependent set 4 problems for the rectangleintersection graph of equal-width rectangles admitPTAS.

2G. Chabert and X. Lorca, On the clique partition of rectangle graphs,Technical Report, 2009.

3A. -A. Mahmood and T. M. Chan, Approximating the piercing number forunit-height rectangles, Proc. CCCG-2005.

4P. K. Agarwal, M. van Kreveld and S. Suri, Label placement by maximumindependent set in rectangles CGTA-1998.


Maximum Independent Set of Rectangle Intersection Graph

Objective: Choose maximum number of mutually non-overlappingrectangles among a set of Rectangles R.

Algorithm:

Draw a vertical line ` at the median of the x-coordinates of thevertical sides of the rectangles.

Split the set R into R1, R2 and R12, whereR12 = set of rectangles stabbed by `;R1 and R2 are the set of rectangles to the left and right of `.

Compute max-indep-set I12 of R12.

Recursively compute max-indep-set I1 and I2.

If |I1|+ |I2| < |I12|, report I = I12 else I = I1 ∪ I2.



Complexity Analysis

In a single level of the recursion hierarchyIdentifying R12 and Computation of I12 needs O(n log n) time.Total size of R1 and R2 = O(n) in the worst case.

Similarly, at each level of recursion, the total number of rectanglesin all the partitions is O(n).

Number of levels of recursion is O(log n).

Thus,Overall running time is O(n log2 n).



Analysis of approximation factor

Approximation factor = |I∗|log n .

Proof: For n = 2, result is obviously true.Let it be true for n ≤ m.Let I ∗, I ∗1 , I ∗2 , I ∗12 are max. independent set of R, R1, R2 and R12.and

I1, I2 and I12 are computed independent set of R1, R2 and R12.

I12 = I ∗12 = I ∗ ∩ R12. |I1| ≥|I∗1 |

log n/2 ≥|I∗∩R1|log n−1 , |I2| ≥ |I

∗∩R2|log n−1

|I | = max(|I12|, |I1|+ |I2|) ≥ max(|I ∗ ∩ R12|, |I∗∩R1|+|I∗∩R2|

log n−1 )

≥ max(|I ∗ ∩ R12|, |I∗|−(|I∗∩R12|

log n−1 )

If |I ∗ ∩ R12| ≥ |I∗|log n then

Theorem is proved.

Otherwise, surely|I∗|−(|I∗∩R12|

log n−1 ≥ |I∗|log n .







I12 = I ∗12 = I ∗ ∩ R12. |I1| ≥|I∗1 |

log n/2 ≥|I∗∩R1|log n−1 , |I2| ≥ |I

∗∩R2|log n−1


log n−1 )

≥ max(|I ∗ ∩ R12|, |I∗|−(|I∗∩R12|

log n−1 )


Theorem is proved.









I12 = I ∗12 = I ∗ ∩ R12. |I1| ≥|I∗1 |

log n/2 ≥|I∗∩R1|log n−1 , |I2| ≥ |I

∗∩R2|log n−1


log n−1 )

≥ max(|I ∗ ∩ R12|, |I∗|−(|I∗∩R12|

log n−1 )


Theorem is proved.









I12 = I ∗12 = I ∗ ∩ R12. |I1| ≥|I∗1 |

log n/2 ≥|I∗∩R1|log n−1 , |I2| ≥ |I

∗∩R2|log n−1


log n−1 )

≥ max(|I ∗ ∩ R12|, |I∗|−(|I∗∩R12|

log n−1 )


Theorem is proved.





Special Case: All rectangles are of same height δ

2-factor Approximation Algorithm:

Split the region into horizontal strips of width δ by drawinghorizontal lines `1, `2, . . . , `k .

Each rectangle will be stabbed by exactly one horizontal line.

For each stabbing line ì , optimally compute the maximumindependent set of all the rectangles stabbed by ì .























Analysis

Compute ISodd = union of the independent sets correspondingto the odd numbered lines.Compute ISeven = union of the independent setscorresponding to the even numbered lines.If |ISodd | < |ISeven| then IS = ISeven else IS = ISodd .

Time complexity: O(n log n)

Approximation Factor: max(|ISodd |, |ISeven| ≥ |I ∗|/2

Application: Map Labelling


Analysis

Compute ISodd = union of the independent sets correspondingto the odd numbered lines.Compute ISeven = union of the independent setscorresponding to the even numbered lines.

If |ISodd | < |ISeven| then IS = ISeven else IS = ISodd .





Analysis






Analysis






Analysis






Polynomial Time Approximation Scheme (PTAS)

Objective:

Given an integer k , find an algorithm for finding an independentset whose size is at least (1 + 1

k )× |I ∗|.

Algorithm:

Partition the plane by horizontal linesat unit interval.

Let Rki be the set of rectangles

intersecting intersected by any one ofthe lines {ì , ì+1, . . . , ì+k−1}.

Define Gj = R1j−1

⋂i≥0 Rk

i(k+1)+j .

Optimally find the largest set ofdisjoint rectangles among those in Gj .

Finally, choose the largest amongthese k independent sets.

`1

`2

`3

`4

`5

`6

`7

`8

`9

`10

R11

R33

R37



Objective:


k )× |I ∗|.

Algorithm:




Define Gj = R1j−1

⋂i≥0 Rk

i(k+1)+j .



`1

`2

`3

`4

`5

`6

`7

`8

`9

`10

R11

R33

R37



Objective:


k )× |I ∗|.

Algorithm:




Define Gj = R1j−1

⋂i≥0 Rk

i(k+1)+j .



`1

`2

`3

`4

`5

`6

`7

`8

`9

`10

R11

R33

R37



Objective:


k )× |I ∗|.

Algorithm:




Define Gj = R1j−1

⋂i≥0 Rk

i(k+1)+j .



`1

`2

`3

`4

`5

`6

`7

`8

`9

`10

R11

R33

R37



Objective:


k )× |I ∗|.

Algorithm:




Define Gj = R1j−1

⋂i≥0 Rk

i(k+1)+j .



`1

`2

`3

`4

`5

`6

`7

`8

`9

`10

R11

R33

R37



Objective:


k )× |I ∗|.

Algorithm:




Define Gj = R1j−1

⋂i≥0 Rk

i(k+1)+j .



`1

`2

`3

`4

`5

`6

`7

`8

`9

`10

R11

R33

R37


Algorithm

Key Observation: No rectangle of Rk1 intersects a rectangle in

Rk+2k . The line `k+1 separates these subgroups.

We use the dynamic programming based routine5. for computingthe largest independent set of rectangles intersected by kconsecutive horizontal lines

5D.S. Hochbaum, W. Maas, Approximation schemes for covering andpacking problems in image processing and VLSI, JACM, 1985.


Analysis

Time Complexity: O(n log n + n2k−1).

Approximation Ratio:

A group is formed by deleting all the rectangles that intersectevery (k + 1)-th. line.

The union of all rectangles in R \ Gj is intersected by at mostd m

k+1e lines.

We compute maximum independent set for each Gj andchoose the largest one.

By pegion hole principle, it can miss at most OPTk+1 rectangles.

Thus we have an (1 + 1k ) factor approximation.


Analysis





k+1e lines.





Analysis





k+1e lines.





Analysis





k+1e lines.





Analysis





k+1e lines.





Analysis





k+1e lines.





Minimum Clique Cover of Rectangle Intersection Graphs

Objective:

Choose minimum number of points on the plane such that eachrectangle contains at least one of those points

These points will be referred as piercing points

Application:

Given a set of points in 2D and a rectangle of specified size, reportthe minimum number of rectangles required to cover all the points.


A 2-approximation algorithm

We consider only the special case where rectangles are of sameheight, say δ.

Algorithm:



For each stabbing line ì , optimally compute the minimumclique cover of all the rectangles Li stabbed by ì .




Algorithm:







Algorithm:







Algorithm:





A 2-approximation algorithm (contd.)

Theorem

The clique cover with all these piercing points is of size 2 ∗ |OPT |,where |OPT | is the size of the optimal clique cover.

Proof:

Let Lodd = {`1, `3, . . .} (* odd numbered stabbing lines *)OPTodd = Piercing points chosen for the rectangles stabbedby the lines in Lodd .

The piercing points chosen for the rectangles stabbed byì ∈ Lodd do not lie in any of the rectangles stabbed by thelines Lodd \ Li .

Thus, OPTodd can be computed in polynomial time.

|OPTodd | ≤ |OPT |, and similarly |OPTeven ≤ |OPT |.

Hence, |OPTodd |+ |OPTeven| ≤ 2 ∗ |OPT |.



Theorem


Proof:








Theorem


Proof:








Theorem


Proof:








Theorem


Proof:







PTAS for Minimum Clique Cover

A PTAS is available for the minimum clique cover problem 6.

Result

For a set of n equal width rectangles and a constant ε, a(1 + ε)-factor approximation of the minimum clique cover can beobtained in O(n1/ε2) time.

6A. -A. Mahmood and T. M. Chan, Approximating the piercing number forunit-height rectangles, Proc. 17th CCCG, 2005.


Unit Disk Graph

Definition

A graph that can be represented as the intersection graph of a setof circles of same radius is called the unit disk graph.

AA

B

C

D

EF

G

H

IJ

B

C

D E

F

G

H

I J


Cliques in Unit Disk Graph

Let vi , vj , vk ∈ V form a clique in the unit disk graph G = (V ,E )and Ci ,Cj ,Ck are the corresponding circles. Now, Ci ,Cj ,Ck mayor may not have a common region.

In the former case it is called a geometric clique, and in the lattercase it is called a graphical clique.

A

B

C

Geometric Clique

A

B

C

Graphical Clique


Largest Clique in Unit Disk Graph

Let A and B be a pair of points such that dist(A,B) ≤ 1.

RAB ⇒ Intersection of two closed disks of radius dist(A,B), andcentered at A and B respectively.

HAB ⇒ RAB⋂

V

Observations:

If A and B are the maximally distantpoints in a set V ′, then V ′ ⊆ HAB .

If C is the vertex set of a maximumsized clique in G then C ⊆ HAB forsome A,B ∈ V with dist(A,B) ≤ 1.

Thus, we inspect all possible pair of points(A,B) with dist(A,B) ≤ 1 to identify apair (a, b) such that Hab contains themaximum clique.

R0AB

R1AB


Largest Clique in Unit Disk Graph

Partition RAB into R0AB and R1

AB by a straight line joining A andB.

Define H0AB = R0

AB ∩ V , and H1AB = R1

AB ∩ V .

Observation: If two points α, β ∈ H0AB (resp. H1

AB), thendist(X ,Y ) ≤ 1. Thus, (α, β) is an edge in G .

Result: The subgraph of G induced by the points in HAB is thecomplement of a bipartite graph.

Computing maximum clique in G is equivalent to computing max.indep. set in G .

Result: The maximum independent set of a bipartite graph with nvertices can be found in O(n2.5) time.

Overall Time Complexity: O(n4.5).


Largest Geometric Clique in Unit Disk Graph

The largest geometric clique in an unit disk graph represents anarea on the plane where maximum number of unit disks overlap.

Applications: Given a set of points and a circle, position the circleon the plane to cover maximum number of points.

Observations:

Each clique region is adjacent to the point of intersection of apair of circles.

The number of intersection points is O(n2) in the worst case.

A trivial O(n3) time algorithm is to consider each intersectionpoint, and inspect how many points are inside that circle.

The best known algorithm runs in O(n2) time7.7S. Cabello, J. Miguel Diaz-Banez, S. Langerman, C. Seara and I. Ventura,

Facility location problems in the plane based on reverse nearest neighborqueries, EJOR-2009


Minimum Clique Cover in Unit Disk Graph

The Problem: Given an unit disk graph, choose the minimumnumber of cliques to cover all the nodes of the graph.

Geometric version: Given a set of circles, choose minimumnumber of points on the floor such that each circle contains atleast one of the chosen points.

Computational hardness: NP-complete 8,

8B. N. Clark, C. J. Colbourn, and D. S. Johnson, Unit disk graphs, DiscreteMathematics,1990.


A 3-Approximation Algorithm

Observation: If the centers of the unit disks lie in a strip of width√3, then the corresponding unit disk graph is a co-comparability

graph9.

Algorithm:

Split the entire plane is divided into strips of width√

3.

Compute the minimum clique cover of the unit disk graphwith disks in each strip. This can be done in polynomial time.

Output all the cliques obtained in all the strips.

Analysis of the approximation factor: Since an unit disk canspan in at most 3 strips, the approximation factor of the algorithmis 3.

9M. R. Cereoli, L. Faria, T. O. Ferreira and F. Protti, On minimum cliquepartition and maximum independent set on unit disk graphs and penny graphs:complexity and approximation, Electronics Notes on Discrete Mathematics,2004


Maximum Independent Set in Unit Disk Graph

The Problem: Given a set S of unit disks, find a subset S∗ of Smaximum cardinality such that the members in S∗ are mutuallynon-overlapping.

Computational hardness: NP-complete 10,

A trivial 3-approximation algorithm:

Choose the left-most disk C0.

Choose C0 in the maximum independent set. This prohibits usnot to choose at most 3 non-intersecting disks, since

the number of mutually non-intersecting disks intersecting C0

is at most 3.

Remove C0 and all the disks overlapping on it from S

Repeat the same steps if the set S is not empty.

10B. N. Clark, C. J. Colbourn, and D. S. Johnson, Unit disk graphs, DiscreteMathematics,1990.


A 2-approximation algorithm for the maximumindependent set for unit disk graph

Divide the plane into strips of width 2. Then an unit disk canspan on at most 2 strips.

We optimally compute the maximum independent set of unitdisks inside a strip as follows:

Divide the strip into squares ofside 2 by drawing vertical lines.

Consider a pair of vertical lines`1 and `2, and a pair of disksintersecting to `1 and `2.

C1

C2

C3

C4

C1

C2

C3

C4

C1

C2C3

C1

C2

C3

C4

C12 C34

C1

C2

C3

C4

C12

C1

C2

C3



For each strip optimally compute the maximum independentset using the following algorithm.

Take the union of the max-indep-sets of either the odd stripsor the even strips (which one is larger).



Max-Indep-Set in a strip:

Construct a graph Gi = (Vi ,Ei ) with the set of circles Ci in astrip i as follows:

Vi : {c ∈ Ci} ∪ {(cα, cβ)| cα and cβ are intersected by a verticalline and they are mutually non-intersecting}Node weight - 1 or 2 depending on its number of circles.

Ei : {(vα, vβ)|vα, vβ ∈ Vi , vα and vβ correspond to two differentvertical line, and the circles of node vα does not intersect withthe circles of node vβ}.



Result

The maximum independent set of the circles in strip i correspondsto the maximum weight path in Gi .

Time Complexity: O(n4)

Reason:

Time complexity of the maximum weight path in a directedacyclic graph - O(m), m = the number of edges in the graph.

Time complexity O(n4) = worst case number of edges in Gi .


Discrete Piercing Set for Unit Disks

Problem

We are given a set of points P, where each point corresponds toan unit disk centered at that point. The objective is to chooseminimum number of points in P such that each disk contains atleast one chosen point.


A constant factor algorithm

Partition the plane into a grid whose each cell is of size1√2× 1√

2

Since the maximum distance between any two points in a gridcell is less than or equal to 1, we can pierce all the diskscentered at points of P in a particular cell by choosing anyone member p ∈ P lying in that cell.

It may cover point(s) in the other cell(s).

But, we show that a disk centered at a point p ∈ P inside agrid cell may cover (some or all) points in at most 14 othergrid cells.




2







2





Algorithm

Consider a 5× 5 grid structure.

The length of its each side of asingle cell is 1√

2× 1√

2.

The cells are numbered as1, 2, . . . , 25.

The cell 13 is split into four parts,namely A, B, C and D.

1 2 3 4 5

6 7 8 9 10

11 12 14 15

16 17 18 19 20

21 22 23 24 25

A B

C D

Observation - 1

A disk of radius 1 centered at any point in sub-cell A maycover (some or all) points in only 15 cells, numbered 2, 3, 4,6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18 and 19.

The same fact can be observed for the sub-cells B, C and D.


Algorithm

Consider a 5× 5 grid structure.

The length of its each side of asingle cell is 1√

2× 1√

2.

The cells are numbered as1, 2, . . . , 25.

The cell 13 is split into four parts,namely A, B, C and D.

1 2 3 4 5

6 7 8 9 10

11 12 14 15

16 17 18 19 20

21 22 23 24 25

A B

C D

Observation - 1

A disk of radius 1 centered at any point in sub-cell A maycover (some or all) points in only 15 cells, numbered 2, 3, 4,6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18 and 19.

The same fact can be observed for the sub-cells B, C and D.


More Detailed Observation

Theorem

A single disk centered at a point inside a cell can not cover pointsin more than 14 cells simultaneously.

a single disk centered at a point p ∈ Acan not cover points in cell number 4and 16 simultaneously.

Reason: Let u and v be the bottom-left andtop-right corners of the cells 4 and 16respectively.

Thus, dist(u, v) = 2,

Let p be a point properly inside cell A.Therefore, dist(u, p) + dist(p, v) > 2.

1 2 3 4 5

6 7 8 9 10

11 12 14 15

16 17 18 19 20

21 22 23 24 25

A B

C D



Theorem

A single disk centered at a point inside a cell can not cover pointsin more than 14 cells simultaneously.

a single disk centered at a point p ∈ Acan not cover points in cell number 4and 16 simultaneously.

Reason: Let u and v be the bottom-left andtop-right corners of the cells 4 and 16respectively.

Thus, dist(u, v) = 2,

Let p be a point properly inside cell A.Therefore, dist(u, p) + dist(p, v) > 2.

1 2 3 4 5

6 7 8 9 10

11 12 14 15

16 17 18 19 20

21 22 23 24 25

A B

C D



Similarly, it can be shown that

a point p ∈ B can cover (some or all) points in cellsnumbered 2, 3, 4. 7, 8, 9, 10, 12, 13, 14, 15, 17, 18, 19 and20, but it can not cover a point in cell 2 and a point in cell 20simultaneously.

a point p ∈ C can cover (some or all) points in cellsnumbered 6, 7, 8, 9, 11, 12, 13, 14, 16, 17, 18, 19, 22, 23 and24, but it can not cover a point in cell 6 and a point in cell 24simultaneously.

a point p ∈ D can cover (some or all) points in cellsnumbered 7, 8, 9, 10, 12, 13, 14, 15, 17, 18, 19, 20, 22, 23and 24 but it can not cover a point in cell 10 and a point incell 22 simultaneously.


Algorithm

Select one point in each non-empty cell for the piercing.

Approximation Factor - 14

Time Complexity - O(n log n)

We shall not construct the grid explicitly.

We maintain an height balanced binary tree T for storing thenon-empty grid cells.

Each element of T is a tuple (α, β) indicating the indices of anon-empty cell.

For each point pi = (xi , yi ) ∈ P, we compute the indices of the gridcell (α, β);Check whether (α, β) ∈ T ;If not present, then insert (α, β) along with pi .

Finally, we visit T , and draw a disk with center at the point pattached to each cell in T .


Conclusion

Some recent results on the approximation algorithms fordifferent optimization problems on rectangle intersectiongraph and unit disk graphs are studied.

The applications of those results in different practicalproblems are also highlighted.

Several other application specific problems are alwaysevolving, and the research in this direction is becomingimportant every day.

Documents

Approximation Algorithms for Geometric Intersection Graphsssen/geomschool/nandy/GEO...Introduction Rectangle Intersection GraphsUnit Disk GraphDiscrete Piercing Set for Unit Disks