4Construct Minimum Unit Disk Cover in Each Cell Each square with edge length1/√2 can be covered by a unitdisk.Hence, each cell can be coveredBy at most disks.1/√2Suppose a cell contains ni points.Then there are ni(ni-1) possiblepositions for each disk.Minimum cover can be computedIn time ni2O(a )
5Solution 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 cellscontaining n1, …, nk points, respectively.Then computing S(x) takes timen n ··· + nk < n2222O(a )O(a )O(a )O(a )
6Approximation Algorithm For x=0, -2, …, -(a-2), compute S(x).Choose minimum one from S(0), S(-2), …, S(-a+2).
7Analysis 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.
8Added Disks So, we have a 4-approximation. Count twice Count four times2So, we have a 4-approximation.
11Estimate # of added disks In vertical strips,each disk appearsonce.
12Estimate # of added disks In horizontal strips,each diskappears once.
13Estimate # 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 addedonly 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.
14Performance Ratio P.R. < 1 + 6/a < 1 + ε when we choose a = 6 ⌠1/ε .O(1/ε )2Running time is n.
16Dominating 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.
17Connected 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.
18Existence of 4-approximation There exists (1+ε)-approximation for minimumdominating set in unit disk graph.2. We can reduce one connected component withat most two nodes.Therefore, there exists a 3(1+ε)-approximation formcds.
19PartitionBut, how do we combine solutionsin each cell together?
20Add extra vertices around the Boundary area central areaBoundary areah+1h
21Why overlapping?cds for Gcds for eachconnectedcomponent1
22Constructing a PTAS In each cell, construct MCDS for each conn. component in the central area.2. Find a 4-approximation D of MCDS of thewhole graph, and add Dbound to the solution
23Step 1 is a restriction:In each central area of a cell e, the feasiblesolution C[e] must satisfy:Each conn. component H of G[e] is dominatedby a single conn. component of C[e].
24The 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.
252. 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.
26MCDS (time) In a square of edge length , any node can dominate every node in the square.Therefore, minimum dominating set has sizeat mosta
272. The total size of MCDSs for connected components in a central area is at mosta
29MCDS (size) Modify a MCDS for G into MCDSs in each cell. D*: MCDS for GD*[e]: MCDS in a cell eD*[e] may not satisfy the restriction; i.e., D*[e] may contain k > 1 components that are in the same component of G[e].
30Estimate P.R. For an MCDS D*, modify it as follows: (1) In each central area, connect allconn. components of D*[e] that are inthe 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.
31Charge the extra vertices to a boundary point Rule 1: Each componentIs charged at most twice.chargingCharged toLeft 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 outsidethe central area.31
32Multiple Charges How many possible charges for each Boundary node? How many componentscan each node beadjacent to?
333 How many independent points can be packed in a half disk with radius 1?3>11
34Each 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.
35Shifting Shifting the partition with distance 1. Each vertex can appear in theboundary area ofat most 4(h+1)partitions.
36Extra nodes in a fixed partition P(a) Each boundary point of D* may be charged 6 timesEach 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*|
37Charging and ShiftingSet h=3p.r.= 1+40(h+1)/aO(a )2Time=n3
38Weighted 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?
39Dominating 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?