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 (x,x) Partition P(x)

Construct Minimum Unit 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 ni points. 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, …, nk points, respectively. Then computing S(x) takes time n1 + n2 + ··· + nk < n 2 2 2 2 O(a ) O(a ) O(a ) O(a )

Approximation Algorithm For x=0, -2, …, -(a-2), compute S(x). Choose minimum one from S(0), S(-2), …, S(-a+2).

Analysis 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.

Added Disks So, we have a 4-approximation. Count twice Count four times 2 So, we have a 4-approximation.

Shifting 2

Estimate # of added disks Shifting

Estimate # of added disks In vertical strips, each disk appears once.

Estimate # of added disks In horizontal strips, each disk appears once.

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 (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.

Performance Ratio P.R. < 1 + 6/a < 1 + ε when we choose a = 6 ⌠1/ε . O(1/ε ) 2 Running time is n.

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

Existence of 4-approximation 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.

Partition But, how do we combine solutions in each cell together?

Add extra vertices around the Boundary area central area Boundary area h+1 h

Why overlapping? cds for G cds for each connected component 1

Constructing a PTAS 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

Step 1 is a restriction: 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.

MCDS (time) In a square of edge length , any node can dominate every node in the square. Therefore, minimum dominating set has size at most . a

2. The total size of MCDSs for connected components in a central area is at most . a

MCDS (size) 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].

Estimate P.R. 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 Dbound to it. Use Charging Method to count the extra from (1). Use Shifting Technique to reduce it.

Charge the extra vertices to a boundary point Rule 1: Each component Is charged at most twice. charging Charged to Left is central area, right is outside central. Charging the two nodes to the node just outside the central area. Rule 2: In each component, charge to the point just outside the central area. 31

Multiple Charges How many possible charges for each Boundary node? How many components can each node be adjacent to?

3 How many independent points can be packed in a half disk with radius 1? 3 >1 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.

Shifting 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 partition P(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*|

Charging and Shifting Set h=3 p.r.= 1+40(h+1)/a O(a ) 2 Time=n 3

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?

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?