chapter 4 partition

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Chapter 4 Partition (1) Shifting Ding-Zhu Du

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Chapter 4 Partition. (1) Shifting. Ding-Zhu Du. Disk Covering. Given a set of n points in the Euclidean plane, find the minimum number of unit disks to cover the n given points. a. (x,x). Partition P(x). Construct Minimum Unit Disk Cover in Each Cell. Each square with edge length - PowerPoint PPT Presentation

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Page 1: Chapter 4 Partition

Chapter 4 Partition

(1) Shifting

Ding-Zhu Du

Page 2: Chapter 4 Partition

Disk Covering

• Given a set of n points in the Euclidean plane, find the minimum number of unit disks to cover the n given points.

Page 3: Chapter 4 Partition

(x,x)

Partition P(x)

a

Page 4: Chapter 4 Partition

Construct Minimum Unit Disk Cover in Each Cell

1/√2

Each square with edge length1/√2 can be covered by a unitdisk.Hence, each cell can be coveredBy at most disks.

Suppose a cell contains ni points.Then there are ni(ni-1) possiblepositions for each disk.

Minimum cover can be computed In time ni

O(a )2

22a

Page 5: Chapter 4 Partition

Solution S(x) associated with P(x)

For each cell, construct minimum cover.S(x) is the union of those minimum covers.

Suppose n points are distributed into k cells containing n1, …, nk points, respectively.Then computing S(x) takes time

n1 + n2 + ··· + nk < nO(a ) O(a ) O(a ) O(a )

2 2 2 2

Page 6: Chapter 4 Partition

Approximation Algorithm

For x=0, -2, …, -(a-2), compute S(x).

Choose minimum one from S(0), S(-2), …, S(-a+2).

Page 7: Chapter 4 Partition

Analysis

• Consider a minimum cover.

• Modify it to satisfy the restriction, i.e.,

a union of disk covers each for a cell.

• To do such a modification, we need to add some disks and estimate how many added disks.

Page 8: Chapter 4 Partition

Added DisksCount twice

Count four times

2

Page 9: Chapter 4 Partition

2

Shifting

Page 10: Chapter 4 Partition

Estimate # of added disks

Shifting

Page 11: Chapter 4 Partition

Estimate # of added disks

Vertical strips

Each disk appearsonce.

Page 12: Chapter 4 Partition

Estimate # of added disks

Horizontal strips

Each disk appears once.

Page 13: Chapter 4 Partition

Estimate # of added disks

# of added disks for P(0)

+ # of added disks for P(-2)+ ···+ # of added disks for P(-a+2)

< 3 opt

where opt is # of disk in a minimum cover.

There is a x such that # of added disks for P(x) < (6/a) opt.

Page 14: Chapter 4 Partition

Performance Ratio

P.R. < 1 + 6/a < 1 + ε when we choose a = 6 ⌠1/ε .

Running time is n.O(1/ε )2

Page 15: Chapter 4 Partition

Unit disk graph

< 1

Page 16: Chapter 4 Partition

Dominating set in unit disk graph

• Given a unit disk graph, find a dominating set with the minimum cardinality.

• Theorem This problem has PTAS.

Page 17: Chapter 4 Partition

Connected Dominating Set in Unit Disk Graph

• Given a unit disk graph G, find a minimum connected dominating set in G.

Theorem There is a PTAS for connected dominating set in unit disk graph.

Page 18: Chapter 4 Partition
Page 19: Chapter 4 Partition

Boundary area

central area

h

h+1

Page 20: Chapter 4 Partition

Why overlapping?

cds for G

cds for eachconnectedcomponent 1

Page 21: Chapter 4 Partition

1. In each cell, construct MCDS for each connected component in the inner area.

Construct PTAS

2. Connect those minimum connected dominating setswith a part of 8-approximation lying in boundary area.

For each partition P(a,a), construct C(a) as follows:

Choose smallest C(a) for a = 0, h+1, 2(h+1), ….

Page 22: Chapter 4 Partition

Existence of 8-approximation

1. There exists (1+ε)-approximation for minimum dominating set in unit disk graph.

2. We can reduce one connected component with two nodes.

Therefore, there exists 3(1+ε)-approximation for mcds.

Page 23: Chapter 4 Partition

8-approximation

1. A maximal independent set has size at most 4 mcds +1.

2. There exists a maximal independent set, connecting it into cds need at most 4mcds nodes.

Page 24: Chapter 4 Partition

MCDS (Time)

2/2

2)2(a

1. In a square of edge length , any node can dominate every bode in the square. Therefore, minimum dominating set has size at most .

a

2/2

Page 25: Chapter 4 Partition

MCDS (Time)

2/2

2. The total size of MCDSs for connected components in an inner square area is at most .

a

3)2(3 a

Page 26: Chapter 4 Partition

nnaO

ii

aO

i

i

n

n

aO

)()(

22

)2(

is time total thecells, allOver

.

time takesareainner in the components

connected allfor cell in the MCDSs all

findingThen nodes. cotains cell a Suppose

Page 27: Chapter 4 Partition

MCDS (Size)

• Modify a mcds for G into MCDSs in each cell.

• mcds(G): mcds for G

• mcdscell(inner): MCDS in a cell for connected components in inner area

Page 28: Chapter 4 Partition

Connect & Charge

charge

Page 29: Chapter 4 Partition

Multiple Charge

charge

How many possiblecharges for each node?

How many componentscan each node be adjacent to?

Page 30: Chapter 4 Partition

1. How many independent points can be packed by a disk with radius 1?

1

>1

5!

Page 31: Chapter 4 Partition

Each node can be charged at most 10 times!!!

charges. 10most at receives nodeEach

nodes. on mde be willchanges 10most At

nodes. 2 tocharge a makecomponent Each

.components 5most at connect tocan nodes

kk

kk

Page 32: Chapter 4 Partition

Shifting

3

a/(2(h+1)) = integer

Time=nO(a )2

h=2

dimesion.any in in timeion approximat-)1( )/1( 2 On

Page 33: Chapter 4 Partition

Weighted Dominating Set

• Given a unit disk graph with vertex weight, find a dominating set with minimum total weight.

• Can the partition technique be used for the weighted dominating set problem?

Page 34: Chapter 4 Partition

Dominating Set in Intersection Disk Graph

• An intersection disk graph is given by a set of points (vertices) in the Euclidean plane, each associated with a disk and an edge exists between two points iff two disks associated with them intersects.

• Can the partition technique be used for dominating set in intersection disk graph?

Page 35: Chapter 4 Partition

Thanks, End