Partial Vertex Cover - Rutgers Universitycrab.rutgers.edu/~rajivg/studentTalks/partialVC.pdfPartial...

Partial Vertex Cover

Joshua Wetzel

Department of Computer ScienceRutgers University–Camden

[email protected]

April 2, 2009

Joshua Wetzel Partial Vertex Cover 1 / 48

Vertex Cover

Input:Given G = (V , E)Non-negative weights on vertices

Objective:Find a least-weight collection of vertices such that eachedge in G in incident on at least one vertex in the collection


Vertex Cover: Example

24

15

50

30

12

10

18

6

COST = 97

24

15

50

30

10

18

6

12

COST = 108


Vertex Cover: IP Formulation

xv ← 1 if v is in our cover, 0 otherwise

min∑

v∈V

wvxv

s.t.

xa + xb ≥ 1, ∀ e = (a, b)

xv ∈ {0, 1}, ∀ v ∈ V


LP Relaxation

Integer programs have been shown to be NP-hard

Relax the integrality constraints xv ∈ {0, 1}, ∀ v ∈ V

min∑

v∈V

wvxv

s.t.

xa + xb ≥ 1, ∀ e = (a, b)

xv ≥ 0, ∀ v ∈ V


Constructing the Dual LP

Primal LP:

min∑

v∈V

wvxv

s.t.

xa + xb ≥ 1, ∀ e = (a, b) (ye)

xv ≥ 0, ∀ v ∈ V


Primal LP and Dual LP

Primal LP:

min∑

v∈V

wvxv

s.t.

xa + xb ≥ 1, ∀ e = (a, b)

xv ≥ 0, ∀ v ∈ V

Dual LP:

max∑

e∈E

ye

s.t. ∑

e : e hits v

ye ≤ wv , ∀ v ∈ V

ye ≥ 0, ∀ e ∈ E


Primal-Dual Method

DualFeasible ≤ DualOPT = PrimalOPT

PrimalOPT ≤ OPTIP

Construct the dual LP

Construct an algorithm that manually tightens dualconstraints to obtain a ’maximal’ dual solution


Clarkson’s Algorithm

Inititally all edges are uncovered.While ∃ an uncovered edge in G:

raise ye for all uncovered edges simultaneouslty until avertex, v , becomes full (i.e

∑e:e hits v

ye = wv )

C ← C ∪ {v}any e that touches v is covered.

Return C as our vertex cover


Clarkson’s Algorithm: Example

∑

e:e hits v

ye ≤ wv

24

15

50

30

12

10

18

00

0

00

0

0

0

0

0

6

Raise each ye uniformly until a vertex is full.



∑

e:e hits v

ye ≤ wv

24

15

50

30

12

10

18

22

2

22

2

2

2

2

2

6



∑

e:e hits v

ye ≤ wv

24

15

50

30

12

10

18

22

2

33

3

3

3

3

3

6



∑

e:e hits v

ye ≤ wv

24

15

50

30

12

10

18

22

2

33

3

3

12

12

12

6


Clarkson’s Algorithm: Example∑

e:e hits v

ye ≤ wv

24

15

50

30

12

10

18

22

2

33

3

3

12

19

19

6

COST = 83


Clarkson’s Algorithm: Analysis

24

15

50

30

12

10

18

22

2

33

3

3

12

19

19

6

∑

e:e hits v

ye ≤ wv

Dual Obj. Fn:

max∑

e

ye

Our Cost = wt(red vertices)

≤ 2∑

e hits red

ye

≤ 2∑

e

ye

= 2DFS

≤ 2OPT


Clarkson’s Algorithm: Tight Example

1 1

1

11

1 6

COSTOPT = 6

1 1

1

11

1 61

11

1

1 1

COSTClarkson = 12


Partial Vertex Cover

Input:Graph, G = (V , E)Non-negative integer weights for verticesInteger, k

Objective:Find the least cost set of vertices in G that will cover atleast k edges.


Key Issue

In full vertex cover OPT covers all edges and hence weknow which edges to cover.

In partial vertex cover, we do not know which k edges OPTcovers.

When k is part of the input, the techniques forfull-coverage do not directly apply.


Clarkson’s Algorithm Fails

Input: k = 1

1

1

1

1

1

1

5

COSTOPT = 1

1

1

1

5/6

5/6

5/65/6

5/6

5/6

5

1

1

1

COSTClarkson = 5


Related Work

Bshouty & Burroughs 1998: solve the LP, modify it andthen round the modified solution. 2-approximation forpartial vertex cover.

Hochbaum 1998: 2-approximation for partial vertex cover.

Bar-Yehuda 1999: “local ratio” method, 2-approximationfor partial vertex cover.

Mestre 2005: 2-approximation primal-dual algorithm forpartial vertex cover with improved running time


Vertex Cover: IP Formulation

xv ← 1 if v is in our cover, 0 otherwise

min∑

v∈V

wvxv

s.t.

xa + xb ≥ 1, ∀ e = (a, b)

xv ∈ {0, 1}, ∀ v ∈ V


Partial Vertex Cover: IP formulation

xv ← 1 if vertex v is chosen in the cover, 0 otherwise.ye ← 1 if edge e is not covered, 0 otherwise.

min∑

v∈V

wvxv

s.t.

ye + xa + xb ≥ 1,∀e = (a, b)∑

e∈E

ye ≤ m − k

xv ∈ {0, 1}, ∀v ∈ V

ye ∈ {0, 1}, ∀e ∈ E


Patial Vertex Cover: LP formulation

Relax the integrality constraints.

xv ∈ {0, 1} → xv ≥ 0, ∀ v ∈ V

ye ∈ {0, 1} → ye ≥ 0, ∀ e ∈ E


Constructing the Dual LP

min∑

v∈V

wvxv

s.t.

ye + xa + xb ≥ 1, ∀ e = (a, b) (ue)∑

e∈E

ye ≤ (m − k) (z)

xv ≥ 0, ∀ v ∈ V

ye ≥ 0, ∀ e ∈ E



Primal LP:

min∑

v∈V

wvxv

s.t.

ye + xa + xb ≥ 1, ∀e∑

e∈E

ye ≤ m − k

xv ≥ 0, ∀ v ∈ V

ye ≥ 0, ∀ e ∈ E

Dual LP:

max∑

e∈E

ue − z(m − k)

s.t. ∑

e : e hits v

ue ≤ wv ,∀v ∈ V

ue ≤ z, ∀ e ∈ E

ue ≥ 0, ∀ e ∈ E

z ≥ 0


Intuition for Primal-Dual

Input: k = 1

1

1

1

1

1

1

5

COSTOPT = 1

1

1

1

5/6

5/6

5/65/6

5/6

5/6

5

1

1

1

COSTClarkson = 5

In last iteration, Clarkson’s Alg. may choose more edgesthan we need at a very high cost

We must somehow bound the cost of last vertex chosen


Primal-Dual Algorithm

For each vertex, v , in GGuess v to be the heaviest vertex in OPT , called vh

Cv ← {vh}Raise weight of all heavier vertices in G to∞Remove all edges touching vh from Gk ′ ← k − deg(vh)Run Clarkson on this instance until k ′ edges are covered.Cchoices ← Cchoices ∪ CV

Return the lowest cost cover, C


Primal-Dual Algorithm: Example

24

15

50

15

12

10

17

18

Input: G = (V , E ), k = 8



∑

e:e hits v

ue ≤ wv

INF

15

INF

15

100

0

0 0

0

0

0

0

INF

INF

12

Vh

k ′ = 6



∑

e:e hits v

ue ≤ wv

INF

15

INF

15

12

104

4

4 4

4

4

4

4

INF

INF

Vh



∑

e:e hits v

ue ≤ wv

INF

15

INF

15

12

106

4

4 4

6

6

6

6

INF

INF

Vh



∑

e:e hits v

ue ≤ wv

INF

15

INF

15

12

1011

4

4 4

11

6

11

11

INF

INF

Vh



∑

e:e hits v

ue ≤ wv

INF

15

INF

15

12

1011

4

4 4

INF

6

INF

INF

Vh

INF

INF

COST =∞


Primal-Dual Algorithm: Example∑

e:e hits v

ue ≤ wv

INF

INF

INF

INF

10INF

INF

10

INF

INF

INF

INF

12

INF

Vh

COST =∞



∑

e:e hits v

ue ≤ wv

INF

10

4

4

6

11

11

17

4

11

12

INF

INF

15

11

Vh

15

COST = 69



∑

e:e hits v

ue ≤ wv

INF

15

INF

10

4 4

INF

6

INF

INF

INF

INF

12

11

4

Vh

15

COST =∞



∑

e:e hits v

ue ≤ wv

INF

INF

INF

INF

10INF

INF

INFINF

INF

INF

INF

INF

INF

INF

INF

Vh

COST =∞



∑

e:e hits v

ue ≤ wv

INF

15

INF

INF12

3

3 3

12

12

17

18

12

3

Vh

15

COST = 60



∑

e:e hits v

ue ≤ wv

24

15

INF

1012

3

3 3

12

7

3

17

18

12

12

Vh

15

COST = 76


Primal-Dual Algorithm: Example∑

e:e hits v

ue ≤ wv

24

15

15

10

3

3 3

6

6

6

17

18

123

50

Vh

COST = 80



24

15

50

15

12

10

17

18

Heaviest Vertex Cost

15 ∞12 ∞17 6915 ∞10 ∞18 6024 7650 80

Return cover with cost of 60


Analysis: Cost of OPT

Ih: inst. in which we correctly guess heaviest vertex in OPT

24

15

50

15

10

17

18

12

Vh

OPT

INF

15

INF

1012

3

3 3

12

12

17

12

3

15

Ih

OPT = OPT (Ih) + w(vh)


Analysis: Our Cost


24

15

50

15

10

17

18

12

Vh

OPT (Ih) + w(vh) = 45

INF

15

INF

1012

3

3 3

12

12

17

18

12

3

Vh

15

COST (Ih) + w(vh) = 60


Analysis: What is z?

INF

15

INF

1012

3

3 3

12

12

17

18

12

3

Vh

Vl

15

Ih

INF

15

INF

10Z

3

3 3

Z

Z

17

18

12

3

VlVh

15

ue ≤ z, ∀ e

z = 12

Num of edges for which ue = z is at least m′ − k ′


Analysis: Our Cost

INF

15

INF

10Z

3

3 3

Z

Z

17

18

12

3

VlVh

15

Ih


Our Cost ≤ Cost(Ih) + w(vh)

= w(RedVert) + w(vl) + w(vh)

≤ w(RedVert) + 2w(vh)

≤ 2∑

e hits red

ue + 2w(vh)

≤ 2[∑

e

ue − z(m′ − k ′)] + 2w(vh)

≤ 2DFS(Ih) + 2w(vh)

≤ 2OPT (Ih) + 2w(vh)

≤ 2OPT



Primal LP:

min∑

v∈V

wvxv

s.t.

ye + xa + xb ≥ 1, ∀e∑

e∈E

ye ≤ m − k

xv ≥ 0, ∀ v ∈ V

ye ≥ 0, ∀ e ∈ E

Dual LP:

max∑

e∈E

ue − z(m − k)

s.t. ∑

e : e hits v

ue ≤ wv ,∀v ∈ V

ue ≤ z, ∀ e ∈ E

ue ≥ 0, ∀ e ∈ E

z ≥ 0


Reference

K. L. Clarkson. A modification of the greedy algorithm forthe vertex cover. Information Processing Letters 16:23-25,1983.

R. Gandhi, S. Khuller, and A. Srinivasan.Approximation Algorithms for Partial Covering Problems.Journal of Algorithms, 53(1):55-84, October 2004.


Thank You.


Partial Vertex Cover - Rutgers Universitycrab.rutgers.edu/~rajivg/studentTalks/partialVC.pdfPartial...

Documents

Transcript of Partial Vertex Cover - Rutgers Universitycrab.rutgers.edu/~rajivg/studentTalks/partialVC.pdfPartial...