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## PowerPoint Slideshow about ' Chapter 4 Partition' - bikita

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I. Covering and Dominating

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

Disk Cover in Each Cell

Each square with edge length

1/√2 can be covered by a unit

disk.

Hence, each cell can be covered

By at most disks.

1/√2

Suppose a cell contains nipoints.

Then there are ni(ni-1) possible

positions for each disk.

Minimum cover can be computed

In time ni

2

O(a )

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, …, nkpoints, respectively.

Then computing S(x) takes time

n1 + n2 + ··· + nk< n

2

2

2

2

O(a )

O(a )

O(a )

O(a )

For x=0, -2, …, -(a-2), compute S(x).

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

- Consider a minimum cover.
- Modify it to satisfy the restriction, i.e.,
a union of disk covers for all cells.

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

2

Shifting

# of added disks for P(0)

+ # of added disks for P(-2)

+ ···

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

< 3 opt

(each disk can be added

only to one P(a).)

where opt is # of disk in a minimum cover.

There exists an x such that

# of added disks for P(x)<(6/a) opt.

< 1

Dominating set in unit disk graph

- Given a unit disk graph, find a dominating set with the minimum cardinality.
- Theorem This problem has PTAS.
- Note: This is just the unit disk covering problem with the restriction that each disk must be centered at an input point.

Connected Dominating Set in Unit Disk Graph

- Given a unit disk graph G, find a minimum connected dominating set in G.
TheoremThere is a PTAS for connected dominating set in unit disk graph.

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

2. We can reduce one connected component with

at most two nodes.

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

mcds.

- In each cell, construct MCDS for each
conn. component in the central area.

2. Find a 4-approximation D of MCDS of the

whole graph, and add Dbound to the solution

In each central area of a cell e, the feasible

solution C[e] must satisfy:

Each conn. component H of G[e] is dominated

by a single conn. component of C[e].

The resulting set of nodes must be a dominating set.

This set is also connected:

1. Two conn. components of Dbound can be connected in D through a conn. component in a central area A. The end points of these components must be dominated by MCDS of A. So, the two components together with MCDS of A are connected together.

2. Every conn. component C of MCDS of a central area is connected to Dbound.

- A point x in C must be dominated by some point y in D.
- y is connected to a point z in Dbound, with all points in the path lying in central area.
- This path and C are in the same conn. component, and so is dominated by C.
- So, C is connected to z.

- In a square of edge length , any node can
dominate every node in the square.

Therefore, minimum dominating set has size

at most .

a

- Modify a MCDS for G into MCDSs in each cell.
- D*: MCDS for G
- D*[e]: MCDS in a cell e
- D*[e] may not satisfy the restriction; i.e., D*[e] may contain k > 1 components that are in the same component of G[e].

For an MCDS D*, modify it as follows:

(1) In each central area, connect all

conn. components of D*[e] that are in

the same component of G[e].

(2) Add Dboundto it.

Use Charging Method to count the extra from (1).

Use Shifting Technique to reduce it.

a boundary point

Rule 1: Each component

Is charged at most twice.

charging

Charged to

Rule 2: In each component,

charge to the point just outside

the central area.

How many possible

charges for each

Boundary node?

charge

How many components

can each node be

adjacent to?

Each node can be charged at most 6 times!!!

Each node can connect to at most 3 components.

Each component makes at most 2 charges to a node.

Therefore, each node can be charged at most 6 times.

- Shifting the
partition with

distance 1.

- Each vertex can
appear in the

boundary area of

at most 4(h+1)

partitions.

Extra nodes in a fixed partitionP(a)

- Each boundary point of D* may be charged 6 times
- Each boundary point of D is used once.
- Together, we get 6 |D*bound| + |Dbound|
By shifting, the total extra nodes in all partitions:

(6 x 4(h+1) + 4 x 4(h+1)) |D*|

- 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?

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?

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