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Improved Approximation Algorithms for the Spanning Star Forest Problem Prasad Raghavendra Ning Chen C. Thach Nguyen Atri Rudra Gyanit Singh University of Washington Roee Engelberg Technion University

Improved Approximation Algorithms for the Spanning Star Forest Problem Prasad Raghavendra Ning ChenC. Thach Nguyen Atri Rudra Gyanit Singh University of

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Improved Approximation Algorithms for the Spanning Star

Forest Problem

Prasad Raghavendra

Ning Chen C. Thach Nguyen Atri Rudra Gyanit Singh

University of Washington

Roee EngelbergTechnion University

Star Forest

A Star is a tree of diameter 2.

A Star Forest is a forest consisting of only Stars (a bunch of vertex disjoint stars )

Center

Leaves

Number of Edges = 4

Unweighted Star Forest Problem

Input : Undirected graph G

Find a Star Forest with the maximum number of edges

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Number of Edges = 5Equivalently, Maximize the number of leaves.

Star Forest Problem

Applications :• Problem of aligning multiple genomic

sequences. [Nguyen .et. al, SODA2007]

• Comparison of Phylogenic Trees.[Berry-Guillemot-Nicholas et. al 2005]

• Diversity Problem in Automobile Industry.[Agra-Cardoso-Cerferia et. al 2005]

Closely Related to the Dominating Set Problem

Dominating Set

A set of vertices S, such that

“Every vertex not in S is adjacent to a vertex in set S”

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Relation to Dominating Set

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L = Set of Leaves

L = Set of NOT Leaves

Every vertex in L is adjacent to a vertex in L.

L is a Dominating Set

Maximum Star Forest = n – (Minimum Dominating Set)

Our Results

We give a 0.71 approximation algorithm for Unweighted star forest problem.– Improves the 0.6 factor in [Nguyen .et. al, SODA2007] New rounding scheme for Dominating Set that yields

- approximation.

– Meets the best known algorithmby analysis of greedy algorithm [P. Slavik, Journal of Algorithms 1997]

Better than ln n when OPT is larger

Our Results

0.64 approximation for the Node-weighted version.

– Nodes have weights , Maximize the total weight of the leaves in the forest.

Hardness of approximation results for the weighted versions of the problem.

– 31/32 hardness for the node weighted version.– 0.95 hardness for the edge weighted version.

Dominating Set - A Linear Program

Variables : (x1, x2 ,… xn)

xi = 1 if vertex i is in dominating set

= 0 otherwiseConstraints : For every vertex, at least one vertex in its neighbourhood belongs to the dominating set

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x1 + x2 + x3 +x4 ≥ 1

Example :

Rounding Scheme

Add vertex i in to the dominating set independently with probability :

Add any vertices still uncovered, to the dominating set.

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Even if xi = 1, probability that it is included is < 1

Let the LP-OPT = a n∙

Analysis

STEP 1 :Add vertex i in to the dominating set independently with probability :

LP Constraint

AnalysisSTEP 2:Add any vertices still uncovered, to the dominating set.

E[Number of Vertices added in STEP 2] ≤ ne-t

Analysis

Linear Programming OPT = a n∙ Expected Size of Dominating Set = n(1–e-at) + ne-t

Choosing the best value of t,

We get a - approximation for dominating set.

Not Enough

Gives good approximation for Star Forest if OPT is closer to nIf OPT is smaller, then gives poorer approximation.

However if OPT is smaller, then there are simple algorithms that give good approximation.

LP OPT = a n∙

Factor =

Simple Tree Algorithm • Pick a spanning tree.• Root the tree at an

arbitrary node.• Divide nodes in to

levels based on distance from root.

• Either the odd or the even levels have at least n/2 nodes.

Make these nodes leaves and other centers.

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A 0.64-Approximation

Tree Algorithm:Finds a Star Forest of size at least n/2So if LP OPT = a n ∙

A 1/2a approximation

LP Algorithm:A approximation

Best of the two algorithms , gives an approximation = 0.64

Getting to 0.7

We design a Combinatorial Algorithm for Unweighted Star Forest that gives

-approximation.

Using this along with LP algorithm gives :

35a

Conclusion

Non linear LP rounding with probability :

1-e-tx

Similar algorithms for Weighted Dominating Set, more generally Weighted Set Cover.

Intuition? Any other applications?

Thank You