Chapter 4 Partition. I. Covering and Dominating. Unit 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. Partition as a restriction. a. (x,x). Partition P(x). Construct Minimum Unit
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I. Covering and Dominating
Disk Cover in Each Cell
Each square with edge length
1/√2 can be covered by a unit
Hence, each cell can be covered
By at most disks.
Suppose a cell contains nipoints.
Then there are ni(ni-1) possible
positions for each disk.
Minimum cover can be computed
In time ni
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
For x=0, -2, …, -(a-2), compute S(x).
Choose minimum one from S(0), S(-2), …,S(-a+2).
a union of disk covers for all cells.
Count four times
So, we have a 4-approximation.
In vertical strips,
each disk appears
In horizontal strips,
# 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.
P.R. <1 + 6/a<1 + ε
when we choose a = 6 ⌠1/ε .
Running time is n.
Connected Dominating Set in Unit Disk Graph
TheoremThere is a PTAS for connected dominating set in unit disk graph.
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
But, how do we combine solutions
in each cell together?
cds for G
cds 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].
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.
dominate every node in the square.
Therefore, minimum dominating set has size
at most .
2. The total size of MCDSs for connected components
in a central area is at most .
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.
Rule 2: In each component,
charge to the point just outside
the central area.
How many possible
charges for each
How many components
can each node be
How many independent points can be packed
in a half disk with radius 1?
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.
appear in the
boundary area of
at most 4(h+1)
By shifting, the total extra nodes in all partitions:
(6 x 4(h+1) + 4 x 4(h+1)) |D*|
Dominating Set in Intersection Disk Graph