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Iterative Rounding and Relaxation
James Davis
Department of Computer ScienceRutgers University–Camden
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
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Outline
1 Introduction: Vertex Cover
2 LP Formulation
3 Algorithm
4 AnalysisBounding CostBounding Degrees
5 Main TheoremLaminar Lemma Proof
6 Improvement
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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
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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′, ∅ >
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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
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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.
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Acknowledgements
Thanks to David Shmoys and David Williamson for letting us use themanuscript of their forthcoming book, “The Design of ApproximationAlgorithms.”
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Thank You!
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