iterative rounding and relaxation - cornell university...delete an xe variable modify (1) constraint...

Post on 07-Mar-2021

3 Views

Category:

Documents

0 Downloads

Preview:

Click to see full reader

TRANSCRIPT

Iterative Rounding and Relaxation

James Davis

Department of Computer ScienceRutgers University–Camden

jamesdav@camden.rutgers.edu

March 11, 2010

James Davis (Rutgers Camden) Iterative Rounding 1 / 58

Iterative Rounding and Relaxation

Ingredients:Linear ProgramTheorem about individual variable values in LP solution

Technique:Solve LPRound some variablesRemove variables, relax constraintsIterate

James Davis (Rutgers Camden) Iterative Rounding 2 / 58

Brief History

Survivable Network DesignJain (1998)

MBDSTGoemans (2006)Singh and Lau (2007, 2008)Bansal, Khandekar, Nagarajan (2008)

James Davis (Rutgers Camden) Iterative Rounding 3 / 58

Outline

1 Introduction: Vertex Cover

2 LP Formulation

3 Algorithm

4 AnalysisBounding CostBounding Degrees

5 Main TheoremLaminar Lemma Proof

6 Improvement

James Davis (Rutgers Camden) Iterative Rounding 4 / 58

Vertex Cover

Input:A graph G = (V ,E)

Non-negative costs on vertices cv

Output:A minimum-cost collection of vertices so that each edge in G isincident on at least one vertex in the collection

James Davis (Rutgers Camden) Iterative Rounding 5 / 58

Vertex Cover

min∑v∈V

cv xv

xu + xv ≥ 1 ∀e = (u, v)

xv ≥ 0 ∀v ∈ V

James Davis (Rutgers Camden) Iterative Rounding 6 / 58

Vertex Cover: Main Theorem

Theorem (Nemhauser-Trotter)

In a basic optimal LP solution, each xv ∈ {12 ,1,0}

Simple 2-appx algorithm:Solve the Vertex Cover LPInclude all vertices with xv 6= 0 in our cover

James Davis (Rutgers Camden) Iterative Rounding 7 / 58

MBDST: Problem Statement

Input:A graph G = (V ,E)

Costs ce ≥ 0 for all e ∈ EA set W ⊆ VDegree bounds bv for all v ∈W

Output:Find a min-cost spanning tree (V ,F ) that doesn’t violate degreebounds.

James Davis (Rutgers Camden) Iterative Rounding 8 / 58

Example

MST

Cost = 3

MBDST

Cost = 7

James Davis (Rutgers Camden) Iterative Rounding 9 / 58

Outline

1 Introduction: Vertex Cover

2 LP Formulation

3 Algorithm

4 AnalysisBounding CostBounding Degrees

5 Main TheoremLaminar Lemma Proof

6 Improvement

James Davis (Rutgers Camden) Iterative Rounding 10 / 58

MBDST Properties

Notation:S: any subset of verticesE(S): edges with both endpoints in SF : set of edges in MBDST

Properties:

Spanning: Exactly |V | − 1 edges in F

Acyclic: For |S| ≥ 2, at most |S| − 1 edges of F in E(S)

Degree Bounds: At most bv edges of F incident on v

James Davis (Rutgers Camden) Iterative Rounding 11 / 58

Integer Program

xe = 1 if e ∈ F and xe = 0 otherwise

min∑e∈E

cexe (Objective)∑e∈E

xe = |V | − 1 (1)∑e∈E(S)

xe ≤ |S| − 1 ∀S ⊂ V , |S| ≥ 2 (2)

∑e∈δ(v)

xe ≤ bv ∀v ∈W (3)

xe ∈ {0,1} ∀e ∈ E

James Davis (Rutgers Camden) Iterative Rounding 12 / 58

Linear Program

xe = 1 if e ∈ F and xe = 0 otherwise

min∑e∈E

cexe (Objective)∑e∈E

xe = |V | − 1 (1)∑e∈E(S)

xe ≤ |S| − 1 ∀S ⊂ V , |S| ≥ 2 (2)

∑e∈δ(v)

xe ≤ bv ∀v ∈W (3)

xe ≥ 0 ∀e ∈ E

James Davis (Rutgers Camden) Iterative Rounding 13 / 58

LP Properties

There are exponentially many constraints (2)

Ellipsoid method

Separation oracleI (1) and (3) are easy to checkI (2) requires work

Skip Oracle

James Davis (Rutgers Camden) Iterative Rounding 14 / 58

Separation Oracle: Flow Network

0

1

1

1

0

James Davis (Rutgers Camden) Iterative Rounding 15 / 58

Separation Oracle: Flow Network

0

1

1

1

0

s

t

James Davis (Rutgers Camden) Iterative Rounding 15 / 58

Separation Oracle: Flow Network

0

1

1

1

0

s

t

James Davis (Rutgers Camden) Iterative Rounding 15 / 58

Separation Oracle: Flow Network

0

1/2

3/2

1

1

1

1/2

1/2

0

s

t

James Davis (Rutgers Camden) Iterative Rounding 15 / 58

Separation Oracle: Flow Network

0

1/2

3/2

1/2

1\2

1/2

1/2

1/2

0

s

t

James Davis (Rutgers Camden) Iterative Rounding 15 / 58

Separation Oracle: Flow Network

0

1/2

3/2

1/2

1\2

1/2

1/2

1/2

0

s

t

11 1

1

James Davis (Rutgers Camden) Iterative Rounding 15 / 58

Separation Oracle:s-t cut

Capacity = 2 + 12 · 2 + 1

2 · 4

James Davis (Rutgers Camden) Iterative Rounding 16 / 58

Separation Oracle

The capacity across S is |V |+ (|S| − 1)−∑

e∈E(S)

xe

The capacity across S is at least |V | iff∑

e∈E(S)

xe ≤ |S| − 1

The max-flow from s to t is |V | iff (2) are satisfied

James Davis (Rutgers Camden) Iterative Rounding 17 / 58

Outline

1 Introduction: Vertex Cover

2 LP Formulation

3 Algorithm

4 AnalysisBounding CostBounding Degrees

5 Main TheoremLaminar Lemma Proof

6 Improvement

James Davis (Rutgers Camden) Iterative Rounding 18 / 58

Main Theorem

x̄ =< x1, x2, . . . , x|E | >: solution to LPSupport(x̄): set of edges s.t. xe > 0

TheoremFor any basic solution x̄ to the linear program either:

1 I ∃v with exactly one incident edge e ∈ Support(x̄)I xe = 1

2 ∃v ∈W with at most 3 edges of Support(x̄) incident on v

Condition 1 identifies a leaf in the treeCondition 2 identifies a vertex with sufficiently small number ofnonzero incident edges

James Davis (Rutgers Camden) Iterative Rounding 19 / 58

Algorithm

F = ∅While |V | > 1

x̄ ← LP solution on < G,W >

Remove all edges e with xe = 0If condition 1 is satisfied by x̄

I Add (u, v) to FI Remove v and (u, v) from GI If u ∈W reduce bu by 1

If condition 2 is satisfied by x̄I Remove v from W

James Davis (Rutgers Camden) Iterative Rounding 20 / 58

Linear Program

min∑e∈E

cexe (Objective)∑e∈E

xe = |V | − 1 (1)∑e∈E(S)

xe ≤ |S| − 1 ∀S ⊂ V , |S| ≥ 2 (2)

∑e∈δ(v)

xe ≤ bv ∀v ∈W (3)

xe ≥ 0 ∀e ∈ E

James Davis (Rutgers Camden) Iterative Rounding 21 / 58

LP Relationships

In each iteration LP is in the same familySame separation oracleThe Main Theorem applies to each LP

If condition 1 is satisfied LP is incrementally modified:Delete an xe variableModify (1) constraintRemove some (2) constraintsModify some (3) constraints

If condition 2 is satisfied LP is incrementally modified:Remove a (3) constraint

James Davis (Rutgers Camden) Iterative Rounding 22 / 58

Outline

1 Introduction: Vertex Cover

2 LP Formulation

3 Algorithm

4 AnalysisBounding CostBounding Degrees

5 Main TheoremLaminar Lemma Proof

6 Improvement

James Davis (Rutgers Camden) Iterative Rounding 23 / 58

Outline

1 Introduction: Vertex Cover

2 LP Formulation

3 Algorithm

4 AnalysisBounding CostBounding Degrees

5 Main TheoremLaminar Lemma Proof

6 Improvement

James Davis (Rutgers Camden) Iterative Rounding 24 / 58

Bounding Cost

TheoremThe tree returned by our algorithm has cost at most LP OPT

e

G'

v1

LP: current lin. prog.LP′: new lin. prog.F ′: MBDST in G′

IH: cost(F ′) ≤ LP′(G′)

cost(F ′) + ce ≤ LP′(G′) + cexe

≤ LP(G′) + cexe

= LP(G)

James Davis (Rutgers Camden) Iterative Rounding 25 / 58

Bounding Cost

LemmaLP(G′) is a feasible solution to LP′(G′)

Changes:

1 −1 on RHS, −1 on LHS2 Remove constraints3 −1 on RHS, −1 on LHS;

Remove constraints

∑e∈E

xe = |V | − 1 (1)∑e∈E(S)

xe ≤ |S| − 1 (2)

∑e∈δ(v)

xe ≤ bv (3)

James Davis (Rutgers Camden) Iterative Rounding 26 / 58

Min-Cost Spanning Trees

Recap:Spanning tree has optimal costDegree bounds?

Implications:

TheoremFor any basic solution x̄ to the linear program either:

1 I ∃v with exactly one incident edge e ∈ Support(x̄)I xe = 1

2 ∃v ∈W with at most 3 edges of Support(x̄) incident on v

James Davis (Rutgers Camden) Iterative Rounding 27 / 58

Min-Cost Spanning Trees

Recap:Spanning tree has optimal costDegree bounds?

Implications:

TheoremFor any basic solution x̄ to the linear program either:

1 I ∃v with exactly one incident edge e ∈ Support(x̄)I xe = 1

2 ∃v ∈W with at most 3 edges of Support(x̄) incident on v

When W = ∅ we have OPT

James Davis (Rutgers Camden) Iterative Rounding 27 / 58

Outline

1 Introduction: Vertex Cover

2 LP Formulation

3 Algorithm

4 AnalysisBounding CostBounding Degrees

5 Main TheoremLaminar Lemma Proof

6 Improvement

James Davis (Rutgers Camden) Iterative Rounding 28 / 58

Degree Bounds

vu 1

bv ,bu ≥ 1bv never violatedbu “adjusted”

Algorithmx̄ ← LP solution on < G,W >Remove e /∈ Support(x̄)If condition 1 is satisfied by x̄

Add (u, v) to F

Remove v and (u, v) from G

If u ∈ W reduce bu by 1

If condition 2 is satisfied by x̄

Remove v from W

James Davis (Rutgers Camden) Iterative Rounding 29 / 58

Degree Bounds

vu 1

bv ,bu ≥ 1bv never violatedbu “adjusted”

Algorithmx̄ ← LP solution on < G,W >Remove e /∈ Support(x̄)If condition 1 is satisfied by x̄

Add (u, v) to F

Remove v and (u, v) from G

If u ∈ W reduce bu by 1

If condition 2 is satisfied by x̄

Remove v from W

James Davis (Rutgers Camden) Iterative Rounding 29 / 58

Degree Bounds

bv ≥ 1All 3 edges may be in Fbv violated by at most 2

Algorithmx̄ ← LP solution on < G,W >Remove e /∈ Support(x̄)If condition 1 is satisfied by x̄

Add (u, v) to F

Remove v and (u, v) from G

If u ∈ W reduce bu by 1

If condition 2 is satisfied by x̄

Remove v from W

James Davis (Rutgers Camden) Iterative Rounding 29 / 58

Degree Bounds

bv ≥ 1All 3 edges may be in Fbv violated by at most 2

Algorithmx̄ ← LP solution on < G,W >Remove e /∈ Support(x̄)If condition 1 is satisfied by x̄

Add (u, v) to F

Remove v and (u, v) from G

If u ∈ W reduce bu by 1

If condition 2 is satisfied by x̄

Remove v from W

James Davis (Rutgers Camden) Iterative Rounding 29 / 58

Degree Bounds

bv ≥ 1All 3 edges may be in Fbv violated by at most 2

Algorithmx̄ ← LP solution on < G,W >Remove e /∈ Support(x̄)If condition 1 is satisfied by x̄

Add (u, v) to F

Remove v and (u, v) from G

If u ∈ W reduce bu by 1

If condition 2 is satisfied by x̄

Remove v from W

James Davis (Rutgers Camden) Iterative Rounding 29 / 58

Analysis Summary

Cost:Spanning tree has optimal cost

Degree Bounds:No degree bound violated by more than 2

Theorem (Goemans)The algorithm for MBDST produces a spanning tree in which thedegree of v is at most bv + 2 for v ∈W and has cost no greater thanOPT

James Davis (Rutgers Camden) Iterative Rounding 30 / 58

Outline

1 Introduction: Vertex Cover

2 LP Formulation

3 Algorithm

4 AnalysisBounding CostBounding Degrees

5 Main TheoremLaminar Lemma Proof

6 Improvement

James Davis (Rutgers Camden) Iterative Rounding 31 / 58

Main Theorem

TheoremFor any basic solution x̄ to the linear program either:

1 I ∃v with exactly one incident edge e ∈ Support(x̄)I xe = 1

2 ∃v ∈W with at most 3 edges of Support(x̄) incident on v

James Davis (Rutgers Camden) Iterative Rounding 32 / 58

Linear Program

min∑e∈E

cexe (Objective)∑e∈E

xe = |V | − 1 (1)∑e∈E(S)

xe ≤ |S| − 1 ∀S ⊂ V , |S| ≥ 2 (2)

∑e∈δ(v)

xe ≤ bv ∀v ∈W (3)

xe ≥ 0 ∀e ∈ E

James Davis (Rutgers Camden) Iterative Rounding 33 / 58

Laminar Lemma

LemmaFor any basic LP solution x̄ there is a Z ⊆W and a collection L ofS ⊆ V where:

1 ∀S ∈ L, S is tight; ∀v ∈ Z, v is tight2 The vectors χE(S) and χδ(v) are independent3 |L|+ |Z | = |Support(x̄)|4 L is laminar

James Davis (Rutgers Camden) Iterative Rounding 34 / 58

Characteristic Vector

χE(S) =< 1,1,1,0,0,0,0 > χδ(v) =< 0,1,1,0,1,0,0 >

James Davis (Rutgers Camden) Iterative Rounding 35 / 58

Laminar Sets

Intersecting Sets

AB

A ∩ B 6= ∅A− B 6= ∅B − A 6= ∅

Laminar Sets

No intersecting sets

James Davis (Rutgers Camden) Iterative Rounding 36 / 58

Laminar Lemma

LemmaFor any basic LP solution x̄ there is a Z ⊆W and a collection L ofS ⊆ V where:

1 ∀S ∈ L, S is tight; ∀v ∈ Z, v is tight2 The vectors χE(S) and χδ(v) are independent3 |L|+ |Z | = |Support(x̄)|4 L is laminar

James Davis (Rutgers Camden) Iterative Rounding 37 / 58

Property of L

LemmaIf all S ∈ L contain at least 2 vertices then |L| ≤ |V | − 1

Use induction on |V |Base case: |V | = 2Induction step

I Shrink smallest set to vertexI Generates L′ and V ′

I L′ is laminarI |L′| = |L| − 1I |V ′| ≤ |V | − 1I |L′| ≤ |V ′| − 1I |L| ≤ |V | − 1

James Davis (Rutgers Camden) Iterative Rounding 38 / 58

Property of L

LemmaIf all S ∈ L contain at least 2 vertices then |L| ≤ |V | − 1

Use induction on |V |Base case: |V | = 2Induction step

I Shrink smallest set to vertexI Generates L′ and V ′

I L′ is laminarI |L′| = |L| − 1I |V ′| ≤ |V | − 1I |L′| ≤ |V ′| − 1I |L| ≤ |V | − 1

James Davis (Rutgers Camden) Iterative Rounding 38 / 58

Property of Support(x̄)

Lemma|Support(x̄)| < |V |+ |W |

Recall property 3 of Laminar Lemma: |L|+ |Z | = |Support(x̄)|

|Support(x̄)| = |L|+ |Z |≤ |L|+ |W |< |V |+ |W | (Previous Lemma)

James Davis (Rutgers Camden) Iterative Rounding 39 / 58

From Laminar Lemma to Main Theorem

TheoremFor any basic solution x̄ to the linear program either:

1 I ∃v with exactly one incident edge e ∈ Support(x̄)I xe = 1

2 ∃v ∈W with at most 3 edges of Support(x̄) incident on v

Suppose Main Theorem wasn’t true:For every v ∈ V there are at least 2 edges incident on itFor every v ∈W there are at least 4 edges incident on it

|Support(x̄)| ≥ 12

(2(|V | − |W |) + 4(|W |))

= |V |+ |W | (Contradiction!)

James Davis (Rutgers Camden) Iterative Rounding 40 / 58

From Laminar Lemma to Main Theorem

TheoremFor any basic solution x̄ to the linear program either:

1 I ∃v with exactly one incident edge e ∈ Support(x̄)I xe = 1

2 ∃v ∈W with at most 3 edges of Support(x̄) incident on v

Suppose Main Theorem wasn’t true:For every v ∈ V there are at least 2 edges incident on itFor every v ∈W there are at least 4 edges incident on it

|Support(x̄)| ≥ 12

(2(|V | − |W |) + 4(|W |))

= |V |+ |W | (Contradiction!)

James Davis (Rutgers Camden) Iterative Rounding 40 / 58

From Laminar Lemma to Main Theorem

TheoremFor any basic solution x̄ to the linear program either:

1 I ∃v with exactly one incident edge e ∈ Support(x̄)I xe = 1

2 ∃v ∈W with at most 3 edges of Support(x̄) incident on v

Suppose Main Theorem wasn’t true:For every v ∈ V there are at least 2 edges incident on itFor every v ∈W there are at least 4 edges incident on it

|Support(x̄)| ≥ 12

(2(|V | − |W |) + 4(|W |))

= |V |+ |W | (Contradiction!)

James Davis (Rutgers Camden) Iterative Rounding 40 / 58

From Laminar Lemma to Main Theorem

TheoremFor any basic solution x̄ to the linear program either:

1 I ∃v with exactly one incident edge e ∈ Support(x̄)I xe = 1

2 ∃v ∈W with at most 3 edges of Support(x̄) incident on v

∑e∈E(S) xe ≤ |V | − 2∑e∈E xe = |V | − 1∑e∈δ(v) xe ≥ 1

xe ≥ 1xe ≤ 1xe = 1

James Davis (Rutgers Camden) Iterative Rounding 41 / 58

From Laminar Lemma to Main Theorem

TheoremFor any basic solution x̄ to the linear program either:

1 I ∃v with exactly one incident edge e ∈ Support(x̄)I xe = 1

2 ∃v ∈W with at most 3 edges of Support(x̄) incident on v

∑e∈E(S) xe ≤ |V | − 2∑e∈E xe = |V | − 1∑e∈δ(v) xe ≥ 1

xe ≥ 1xe ≤ 1xe = 1

James Davis (Rutgers Camden) Iterative Rounding 41 / 58

From Laminar Lemma to Main Theorem

TheoremFor any basic solution x̄ to the linear program either:

1 I ∃v with exactly one incident edge e ∈ Support(x̄)I xe = 1

2 ∃v ∈W with at most 3 edges of Support(x̄) incident on v

S=V-v

v∑

e∈E(S) xe ≤ |V | − 2∑e∈E xe = |V | − 1∑e∈δ(v) xe ≥ 1

xe ≥ 1xe ≤ 1xe = 1

James Davis (Rutgers Camden) Iterative Rounding 41 / 58

From Laminar Lemma to Main Theorem

TheoremFor any basic solution x̄ to the linear program either:

1 I ∃v with exactly one incident edge e ∈ Support(x̄)I xe = 1

2 ∃v ∈W with at most 3 edges of Support(x̄) incident on v

S={u,v}u v

∑e∈E(S) xe ≤ |V | − 2∑e∈E xe = |V | − 1∑e∈δ(v) xe ≥ 1

xe ≥ 1xe ≤ 1xe = 1

James Davis (Rutgers Camden) Iterative Rounding 41 / 58

Outline

1 Introduction: Vertex Cover

2 LP Formulation

3 Algorithm

4 AnalysisBounding CostBounding Degrees

5 Main TheoremLaminar Lemma Proof

6 Improvement

James Davis (Rutgers Camden) Iterative Rounding 42 / 58

Laminar Lemma

LemmaFor any basic LP solution x̄ there is a Z ⊆W and a collection L ofS ⊆ V where:

1 ∀S ∈ L, S is tight; ∀v ∈ Z, v is tight2 The vectors χE(S) and χδ(v) are independent3 |L|+ |Z | = |Support(x̄)|4 L is laminar

James Davis (Rutgers Camden) Iterative Rounding 43 / 58

LP Background

min∑

cixi (Objective)

a11x1 + a12x2 + ...+ a1nxn ≥ b1 (1)a21x1 + a22x2 + ...+ a2nxn ≥ b2 (2)

... = ...

am1x1 + am2x2 + ...+ amnxn ≥ bm (m)xi ≥ 0 (Non-Negative)

James Davis (Rutgers Camden) Iterative Rounding 44 / 58

LP Background

Constraints definehalf-spacesObjective is a hyperplaneSolution always a corner≥ n tight constraintsConstraints lin. ind.

Linear Program

min∑

ci xi

a11x1 + a12x2 + ...+ a1nxn ≥ b1 (1)

a21x1 + a22x2 + ...+ a2nxn ≥ b2 (2)

... = ...

am1x1 + am2x2 + ...+ amnxn ≥ bm (m)

xi ≥ 0

James Davis (Rutgers Camden) Iterative Rounding 45 / 58

LP Background

Constraints definehalf-spacesObjective is a hyperplaneSolution always a corner≥ |E | tight constraintsConstraints lin. ind.

MBDST LP

min∑e∈E

cexe

∑e∈E

xe = |V | − 1 (1)

∑e∈E(S)

xe ≤ |S| − 1 (2)

∑e∈δ(v)

xe ≤ bv (3)

xe ≥ 0

James Davis (Rutgers Camden) Iterative Rounding 45 / 58

Laminar Lemma

LemmaFor any basic LP solution x̄ there is a Z ⊆W and a collection L ofS ⊆ V where:

1 ∀S ∈ L, S is tight; ∀v ∈ Z, v is tight2 The vectors χE(S) and χδ(v) are independent3 |L|+ |Z | = |Support(x̄)|4 L is laminar

James Davis (Rutgers Camden) Iterative Rounding 46 / 58

Laminar Lemma

LemmaFor any basic LP solution x̄ there is a Z ⊆W and a collection L ofS ⊆ V where:

1 ∀S ∈ L, S is tight; ∀v ∈ Z, v is tight2 The vectors χE(S) and χδ(v) are independent3 |L|+ |Z | = |Support(x̄)|4 L is laminar

James Davis (Rutgers Camden) Iterative Rounding 46 / 58

Laminar Lemma Proof

Lemma∑e∈E(S) xe is supermodular∑

e∈E(S)

xe +∑

e∈E(T )

xe ≤∑

e∈E(S∩T )

xe +∑

e∈E(S∪T )

xe

S T

James Davis (Rutgers Camden) Iterative Rounding 47 / 58

Laminar Lemma Proof

LemmaS,T are tight, S and T cross, then S ∩ T ,S ∪ T are tight and

χE(S) + χE(T ) = χE(S∩T ) + χE(S∪T )

(|S| − 1) + (|T | − 1) = (|S ∩ T | − 1) + (|S ∪ T | − 1)

≥∑

E(S∩T )

xe +∑

E(S∪T )

xe (feasibility)

≥∑E(S)

xe +∑E(T )

xe (supermodularity)

Since S and T are tight, these are all equalities

James Davis (Rutgers Camden) Iterative Rounding 48 / 58

Laminar Lemma Proof: Finding L

Lemma∃L that is laminar and Span(T ) ⊆ Span(L), where T contains all tightsets

Let L be a maximal laminar collection of TRecall that χE(S) + χE(T ) = χE(S∩T ) + χE(S∪T )

S T

Span(T ) Span(L)

S (least int. in L) T (int. S)S ∪ T and S ∩ T S ∪ T and S ∩ T

S ∩ T S ∪ TS ∪ T S ∩ T

James Davis (Rutgers Camden) Iterative Rounding 49 / 58

Laminar Lemma Proof: Finding L

Lemma∃L that is laminar and Span(T ) ⊆ Span(L), where T contains all tightsets

Let L be a maximal laminar collection of TRecall that χE(S) + χE(T ) = χE(S∩T ) + χE(S∪T )

S T

Span(T ) Span(L)

S (least int. in L) T (int. S)S ∪ T and S ∩ T S ∪ T and S ∩ T

S ∩ T S ∪ TS ∪ T S ∩ T

James Davis (Rutgers Camden) Iterative Rounding 49 / 58

Laminar Lemma Proof: Finding L

Lemma∃L that is laminar and Span(T ) ⊆ Span(L), where T contains all tightsets

Let L be a maximal laminar collection of TRecall that χE(S) + χE(T ) = χE(S∩T ) + χE(S∪T )

S T

Span(T ) Span(L)

S (least int. in L) T (int. S)S ∪ T and S ∩ T S ∪ T and S ∩ T

S ∩ T S ∪ TS ∪ T S ∩ T

James Davis (Rutgers Camden) Iterative Rounding 49 / 58

Laminar Lemma Proof: Finding L

Lemma∃L that is laminar and Span(T ) ⊆ Span(L), where T contains all tightsets

Let L be a maximal laminar collection of TRecall that χE(S) + χE(T ) = χE(S∩T ) + χE(S∪T )

S T

Span(T ) Span(L)

S (least int. in L) T (int. S)S ∪ T and S ∩ T S ∪ T and S ∩ T

S ∩ T S ∪ TS ∪ T S ∩ T

James Davis (Rutgers Camden) Iterative Rounding 49 / 58

Laminar Lemma Proof: Finding L

Lemma∃L that is laminar and Span(T ) ⊆ Span(L), where T contains all tightsets

Let L be a maximal laminar collection of TRecall that χE(S) + χE(T ) = χE(S∩T ) + χE(S∪T )

T

Span(T ) Span(L)

S (least int. in L) T (int. S)S ∪ T and S ∩ T S ∪ T and S ∩ T

S ∩ T S ∪ TS ∪ T S ∩ T

James Davis (Rutgers Camden) Iterative Rounding 49 / 58

Laminar Lemma Proof: Finding L

Lemma∃L that is laminar and Span(T ) ⊆ Span(L), where T contains all tightsets

Let L be a maximal laminar collection of TRecall that χE(S) + χE(T ) = χE(S∩T ) + χE(S∪T )

T

Span(T ) Span(L)

S (least int. in L) T (int. S)S ∪ T and S ∩ T S ∪ T and S ∩ T

S ∩ T S ∪ TS ∪ T S ∩ T

James Davis (Rutgers Camden) Iterative Rounding 49 / 58

Laminar Lemma Proof: Finding Z

LemmaFor any basic LP solution x̄ there is a Z ⊆W and a collection L ofS ⊆ V where:

1 ∀S ∈ L, S is tight; ∀v ∈ Z, v is tight2 The vectors χE(S) and χδ(v) are independent3 |L|+ |Z | = |Support(x̄)|4 L is laminar

(T ,Y ) spans R|Support(x̄)|

(L,Y ) spans R|Support(x̄)|

To obtain (L,Z ) remove v ∈ Y that are dependent

James Davis (Rutgers Camden) Iterative Rounding 50 / 58

Recap

LP formulationMain TheoremAlgorithmCost no more than OPTDegree bounds violated by at most 2Main Theorem ProofLaminar Lemma Proof

James Davis (Rutgers Camden) Iterative Rounding 51 / 58

Outline

1 Introduction: Vertex Cover

2 LP Formulation

3 Algorithm

4 AnalysisBounding CostBounding Degrees

5 Main TheoremLaminar Lemma Proof

6 Improvement

James Davis (Rutgers Camden) Iterative Rounding 52 / 58

Improved Main Theorem

TheoremIf x̄ is a basic solution to LP where W 6= ∅ then there is a v, s.t.

|δ(v) ∩ Support(x̄)| ≤ bv + 1

James Davis (Rutgers Camden) Iterative Rounding 53 / 58

Algorithm

Phase 1:While W 6= ∅

x̄ ← LP solution on < G,W >

For all xe = 0, remove e from ERemove v from W if there are at most bv + 1 edges of δ(v) inSupport(x̄)

Phase 2:Run algorithm on < G′, ∅ >

James Davis (Rutgers Camden) Iterative Rounding 54 / 58

Analysis

Theorem (Singh and Lau)The improved algorithm for MBDST produces a spanning tree in whichthe degree of v is at most bv + 1 for v ∈W and has cost no greaterthan OPT

James Davis (Rutgers Camden) Iterative Rounding 55 / 58

References

Kamal Jain. A factor 2 approximation algorithm for the generalizedSteiner network problem. Combinatorica, 21:39-60, 2001.

Michel X. Goemans. Minimum bounded-degree spanning trees. FOCS’06

Mohit Singh and Lap Chi Lau. Approximating minimum boundeddegree spanning trees to within one of optimal. STOC ’07.

James Davis (Rutgers Camden) Iterative Rounding 56 / 58

Acknowledgements

Thanks to David Shmoys and David Williamson for letting us use themanuscript of their forthcoming book, “The Design of ApproximationAlgorithms.”

James Davis (Rutgers Camden) Iterative Rounding 57 / 58

Thank You!

Return to Oracle

James Davis (Rutgers Camden) Iterative Rounding 58 / 58

top related