Chapter 6. Relaxation

Chapter 6. Relaxation II. Connected Dominating Set

Connected Dominating Set . . Dominating set Connected dominating set

Notation . With respect to a vertex set C, . Color vertices in C by Black . Color vertices adjacent to a black vertex by Grey Color all other vertices by White

A Greedy Algorithm? What is the potential function? Following the idea on Connected Vertex Cover, we may try g(C) = p(C) +h(C) • p(C)= # of connected components of the subgraph induced by C (set of black vertices) • h(C)= # of white vertices (not dominated by C) Similar to CVC,h(C) is submodular, but – p(C) is not submodular.

In addition, in Min-CDS, g(C) = p(C) + h(C) is not a good potential function?? g(C) = 1 C is a cds But, it is possible that C may not be a connected even if x g(C) = 0 for all x.

Lemma If x g(C) = 0for all x, then (1) C must be a dominating set (no white vertices) (2) if C is not connected, then 2 black components can be connected by 2 grey vertices. Proof (1) G is connected. If there is a white vertex, there must be one that is adjacent to a grey vertex x. Adding x to C does not change p(C) but decreases h(C). (2) If not true: What color?

Two-Stage Greedy Approximation Stage 1. Find C such that - x g(C) = 0 for allx. {Relax CDS to DS with min. p(C)} Stage 2. While p(C) > 0 Find 2 grey vertices connecting 2 black Components, and add them to C. {Modify the solution to make it a CDS}

This is a (3+ ln )-Approximation whereis the max degree of the input graph Proof Suppose the algorithm selects in stage 1. Let A piece is either a white vertex or a black component. A vertex xtouches a piece if x is in it or is adjacent to it. If x touches m pieces, then - x g(C) If x has k white neighbors and m – k black component neighbors, then coloring it black reduces k white vertices and m – k – 1 black components.

Proof (cont'd) By Pigeonhole, where must be < 0 (otherwise – xg(Ct) > 0)

opt t 0 i j

# of vertices used in stage 1 = t # of vertices used in stage 2 <2(opt - t + j) How many pieces left after stage 1? g(Ct) = opt + at< opt – t + j

Actually, we can use a different potential function for a one-stage greedy (1+)(1+ln )-approximation: g(C) = p(C) +q(C) • p(C)= # of connected components of the subgraph induced by C (set of black vertices) • q(C)= # of connected components of (V, D(C)), where D(C) is the set of edges incident on C. Note: – g(C)is not submodular. Need to explore the greedy algorithm more carefully.

CDS in unit disk graphs

CDS in unit ball graphs

Two Stage Algorithm Stage 1. Compute a maximal independence set D. Relax to the problem of Max Independent Set with the following property: For every subset S of D, there is a vertex x adjacent to both S and D – S. Stage 2. Connect D into a connected dominating set. Modification

Stage 1 The Special property is easy to satisfy: Add a white vertex to the independent set only if it has a grey neighbor.

How good is the max independent set as an approximation to MCDS? How many independent points can be contained by a disk with radius 1? 5

How many independent points can be contained by two disks with radius 1 and center distance < 1? 8 (Wu et al, 2006)

How many independent points can be packed Into four disks that one contains centers of other three? < 15 (Yao et al, 2008)

How good is the max independent set as an approximation to MCDS? There exist  and  such that | MIS | < | MCDS | +  Conjecture: = 3,  = 2. We give a simple proof for  = 3.74,  = 5.26

How much area can a CDS of size n cover? Vertex x dominates vertex y if | x – y | < 1. So, the area covered by x and all its neighbors is within distance 3/2 of x. Two disks of radius 3/2 and having centers within distance 1 have a common area of size >

Ax1 = 2/3, q = arccos (1/3) 2 shaded area = (3/2) q/2 area of =

So, the total area covered by a CDS of size n is at most How many independent disks can we put in this area?

Stage 2. Apply greedy algorithm with potential function – p(C), where p(C) is # of connected components of black vertices. Note the special property of D at the end of stage 1: For every connected component of D, there is a vertex x adjacent to D and another component. From this property, the output of stage 2 must be a connected dominating set.

Assume that | MIS | <opt +  Then, the output of stage 2 has size at most which is at most if  = 3.74 and  = 5.26 Proof is similar to that for other greedy algorithms.

CDS in unit ball graphs What are  and ? 1. How many independent points can be packed by a ball with radius 1? 1 >1

2. How many (untouched) unit balls can be packed into a ball with radius 1.5? 0.5 1.5

3. Gregory-Newton Problem (1694) How many unit balls (not touch each other) can kiss a unit ball?

Relationship between problems 1, 2 and 3? 1.5 1 .5

In unit ball graph (Butenko, et al, 2007) 11 12 11

Stage 2 Connect all nodes in an MIS with a spanning tree for unit disk graphs (Wan-Yao) for unit ball graphs (Butenko, 2007)

Multiway Cut Given a graph with weighted edges, and a set of terminals x1, x2, … , xk, find a set of edges of min total weight that cut these terminals from each other. When k = 2, this is the minimum cut problem.

When we remove the cutting edges, the graph has exactly k connected componenets each containing an input terminal: G1, G2, … , Gk. Let Ci denote the set of cutting edges incident upon Gi. Then, Ci is a cut of xi from other terminals.

A relaxation algorithm Relax the problem to a cut of a single terminal from k – 1 terminals. Let Di denote the cut for xi from all other terminals. Modification: combine all (single-terminal) cuts to a multiway cut. Let C be the union of Di 's.

Complexity Min-Cut has a poly-time algorithm Analysis ? The approx solution is a combination of a number of relaxed solutions. How do we analyze the error?

Alternative view This algorithm may be viewed as a restriction algorithm: First, restrict the multiway cut must be the union of single-terminal cuts. Then, the error can be estimated by converting an opt. solution to the restricted solution: Let Gi and Ci be the connected component and the cut obtained from C* for terminal xi. We have w(Di) < w(Ci). Also, each edge can occur In at most two Ci 's. So, this is a 2-approximation.

Lesson: design idea ← relaxation error analysis ← restriction

Chapter 6. Relaxation