On the Max k-Vertex Cover problem - TCS...

IntroductionLiterature

An exact approach combining Branch-and-reduce and Measure-and-conquer

On the Max k-Vertex Cover problem

F. Della Croce, V. Th. Paschos

[email protected]@lamsade.dauphine.fr

F. Della Croce, V. Th. Paschos On the Max k-Vertex Cover problem

Outline

1 Introduction

2 Literature

3 An exact approach combining Branch-and-reduce andMeasure-and-conquer

Introduction

Max k-Vertex problem (also known as partial vertex cover):given a graph G (V ,E ) of order n and a constant k < n, wesearch for k vertices that cover the maximum number ofedges in G .Complexity:NP-hard (straightforward reduction from the minimum vertexcover problem).Notation:We use notation O∗(·) to measure the running time of analgorithm ignoring polynomial factors.

Introduction

Literature

Apparently no polynomial-space exact approaches availableswith complexity lower than the trivial 2n. An exact algorithmwith complexity O∗(n0.792k) was proposed by Cai (ComputerJournal 08).Fixed parameter tractability:

W[1]-hard wrt parameter k (very little hope that an algorithmsolving Max k-Vertex problem with running time O∗(f (k))could ever be devised) - Guo et al., WADS05.in FPT wrt standard parameter l (1.396l - Kneis et al., WG08).in FPT wrt parameter w (w = graph treewidth: 2w - Moser,05).

Literature

An exact approach combining Branch-and-reduce andMeasure-and-conquer (I)

Branch-and-reduce:Search tree algorithm where at each branch the problem size istypically reduced in terms of variables.Measure-and-conquer: (Fomin et al. JACM09):weights are typically added to the variables and such weightsopportunately decrease when the search tree levels increase.Branching scheme: vertex j is either selected or discarded.Remark: with a pure branch-and-reduce approach, noimprovement occurs wrt the trivial O∗(2n).

An exact approach combining Branch-and-reduce andMeasure-and-conquer (II)

Further notation:dj = degree of vertex jN(j) the subset of vertices adjacent to j .αj = total number of vertices adjacent to j that have beendiscarded in the previous levels of the search tree.cj = dj − αj .

Remark: - Branch on j -for each l ∈ N(j),if j is selected, then dl is decreased by one unit as edge (j , l) isalready covered by j ;if j is discarded, then dl is not modified and αl is increased byone unit.

An exact approach combining Branch-and-reduce andMeasure-and-conquer (III)

Measure and conquer:we do not count in the measure the fixed vertices (those thathave been either selected or discarded at an earlier stage of thesearch tree) and we count with a weight wh the free (unfixed)vertices h.

An exact approach combining Branch-and-reduce andMeasure-and-conquer (IV)

Weights of the free vertices:each free vertex h is assigned a weight wh = w[i ] withi = ci = dh − αh and we imposew[0] ≤ w[1] ≤ w[2] ≤ w[3] ≤ . . . ≤ w[cmax] = 1 that is theweights of the vertices are strictly increasing in their cjcoefficients.Branching decisionWe branch on the vertex j with largest coefficient cj = dj − αj(notice that cmax ≤ ∆).

An exact approach combining Branch-and-reduce andMeasure-and-conquer (IV)

Weights of the free vertices:each free vertex h is assigned a weight wh = w[i ] withi = ci = dh − αh and we imposew[0] ≤ w[1] ≤ w[2] ≤ w[3] ≤ . . . ≤ w[cmax] = 1 that is theweights of the vertices are strictly increasing in their cjcoefficients.Branching decisionWe branch on the vertex j with largest coefficient cj = dj − αj(notice that cmax ≤ ∆).

An exact approach combining Branch-and-reduce andMeasure-and-conquer (V)

We get recurrences on the time T (p) required to solveinstances of size p, where the size of an instance is the sum ofthe weights of its vertices.For p = 0, there are only vertices with weight w0 in the graphand, in this case, the problem is immediately solved byselecting the k − γ vertices with largest αj (if γ < k verticeshave been selected so far). Hence, free vertices j with noadjacent free vertices receive weight w[0] = 0.

An exact approach combining Branch-and-reduce andMeasure-and-conquer (V)

We get recurrences on the time T (p) required to solveinstances of size p, where the size of an instance is the sum ofthe weights of its vertices.For p = 0, there are only vertices with weight w0 in the graphand, in this case, the problem is immediately solved byselecting the k − γ vertices with largest αj (if γ < k verticeshave been selected so far). Hence, free vertices j with noadjacent free vertices receive weight w[0] = 0.

The exact algorithm MAXKVC (I)

MAXKVC:Select j such that cj is maximum and branch according to thefollowing exhaustive cases:

1 if cj ≥ 3, then branch on j and either select or discard j ;2 else, cj ≤ 2 and MAXKVC is polynomially solvable (by dynamic

programming - Niedermeier et al. JA03).

Theorem

Algorithm MAXKVC solves the Max k-Vertex Cover problem withrunning time O∗(2

∆−1∆+1 n).

The exact algorithm MAXKVC (I)

MAXKVC:Select j such that cj is maximum and branch according to thefollowing exhaustive cases:

1 if cj ≥ 3, then branch on j and either select or discard j ;2 else, cj ≤ 2 and MAXKVC is polynomially solvable (by dynamic

programming - Niedermeier et al. JA03).

Theorem

Algorithm MAXKVC solves the Max k-Vertex Cover problem withrunning time O∗(2

∆−1∆+1 n).

The exact algorithm MAXKVC (II)

Sketch of proof

Branch on the vertex j with largest cj = cmax ≤ ∆ wherecj ≥ 3 and either we select or discard j .If we select j , vertex j is fixed and cmax vertices (the neighborsof j) decrease degree and coefficient by one unit.If we discard j , vertex j is fixed and cmax vertices (theneighbors of j) decrease their coefficient as their α parameteris increased by one unit.

Sketch of proof

The exact algorithm MAXKVC (III)

The recurrence isT (p) ≤ 2T

(p − w[cmax] −

∑h∈N(j)

(w[ch] − w[ch−1]

))By constraining the weights to satisfy the inequalityw[j] − w[j−1] ≤ w[j−1] − w[j−2], ∀j = 2, . . . , cmax, therecurrence becomes in the worst-caseT (p) ≤ 2T

(p − w[cmax] − cmax

(w[cmax] − w[cmax−1]

As cmax ≤ ∆, where the equality occurs when αj = 0, theabove recurrence becomes, in the worst-case,T (p) ≤ 2T

(p − w[∆] −∆

(w[∆] − w[∆−1]

(p − w[cmax] −

∑h∈N(j)

(w[ch] − w[ch−1]

(p − w[∆] −∆

(w[∆] − w[∆−1]

(p − w[cmax] −

∑h∈N(j)

(w[ch] − w[ch−1]

(p − w[∆] −∆

(w[∆] − w[∆−1]

The exact algorithm MAXKVC (IV)

All together we want to satisfy contemporaneouslythe recurrencesT (p) ≤ 2T (p − w[i ] − i(w[i ] − w[i−1])), ∀i ∈ 3, . . . ,∆ (ascj ≥ 3)

the constraints

w[i ] − w[i−1] ≤ w[i−1] − w[i−2] ∀i = 2, . . . ,∆0 = w[0] ≤ w[1] ≤ w[2] ≤ w[3] ≤ . . . ≤ w[∆−1] ≤ w[∆] = 1

This corresponds to a non linear optimization problem of the form:

min α

α(w[i ]+i(w[i ]−w[i−1])) ≥ 2 ∀i = 3, . . . ,∆w[i ] − w[i−1] ≤ w[i−1] − w[i−2] ∀i = 2, . . . ,∆

0 = w[0] ≤ w[1] ≤ w[2] ≤ w[3] ≤ . . . ≤ w[∆−1] ≤ w[∆] = 1

the constraints

min α

0 = w[0] ≤ w[1] ≤ w[2] ≤ w[3] ≤ . . . ≤ w[∆−1] ≤ w[∆] = 1

the constraints

min α

0 = w[0] ≤ w[1] ≤ w[2] ≤ w[3] ≤ . . . ≤ w[∆−1] ≤ w[∆] = 1

The exact algorithm MAXKVC (V)

By means of a non linear solver, we get the performances of thealgorithm for small values of ∆.

∆ 3 4 5 6 7 8 9O∗(n) 1.414n 1.516n 1.587n 1.641n 1.682n 1.714n 1.741n

For all values of ∆, the complexity corresponds to O∗(2∆−1∆+1 n).

This is not accidental:, by setting:

w[i ] =(i − 1)(∆ + 1)

(i + 1)(∆− 1)∀i = 2, . . . ,∆

w[1] =12w[2], w[0] = 0

the constraints are satisfied and the complexity is O∗(2∆−1∆+1 n).

∆ 3 4 5 6 7 8 9O∗(n) 1.414n 1.516n 1.587n 1.641n 1.682n 1.714n 1.741n

w[i ] =(i − 1)(∆ + 1)

(i + 1)(∆− 1)∀i = 2, . . . ,∆

w[1] =12w[2], w[0] = 0

∆ 3 4 5 6 7 8 9O∗(n) 1.414n 1.516n 1.587n 1.641n 1.682n 1.714n 1.741n

w[i ] =(i − 1)(∆ + 1)

(i + 1)(∆− 1)∀i = 2, . . . ,∆

w[1] =12w[2], w[0] = 0

Tailoring for ∆ = 3

PropositionAlgorithm MAXKVC-3 solves the Max k-Vertex Cover Problem ongraphs with maximum degree 3 with running time O∗(1.3339n).

Parameterization by τ

TheoremMax k-Vertex Cover Problem is solvable with running time O∗(2τ )where τ is the size of the minimum vertex cover.

1 compute a minimum vertex cover V ′ O∗(1.2738τ );

2 for every subset S ′ ⊆ V ′ of cardinality at most k , take thek − |S ′| vertices of V \ V ′ with the largest degrees to V ′ \ S ′;denote by I ′ this latter set (O∗(

∑ki=1(τi

)) < O∗(2τ ));

3 return the best among the sets S ′ ∪ I ′ so-computed (i.e., theset that covers the maximum of edges).

∑ki=1(τi

)) < O∗(2τ ));

∑ki=1(τi

)) < O∗(2τ ));

∑ki=1(τi

)) < O∗(2τ ));

Further results

Max k-Vertex Cover can be solved in time O∗(max{γτ , ck}),for two constants γ < 2 and c > 4, and with polynomialspace.

Max k-Vertex Cover can be solved to optimality in O∗(∆k)and polynomial space.

Max k-Vertex Cover can be approximated within ratio 1− εand with running time:

O∗(n(1+2

√1−3ε)(ωk)/9

),O∗

1 + 2√1− 3ε

)}and polynomial space.

Further results

O∗(n(1+2

√1−3ε)(ωk)/9

),O∗

1 + 2√1− 3ε

Further results

O∗(n(1+2

√1−3ε)(ωk)/9

),O∗

1 + 2√1− 3ε

Final issue

Do you know anything about Min k-Vertex Cover?

On the Max k-Vertex Cover problem - TCS...

Documents

Transcript of On the Max k-Vertex Cover problem - TCS...