Integrality Gaps of Bidirected Cut and Directed Component ... · Bidirected Cut and Directed...
Transcript of Integrality Gaps of Bidirected Cut and Directed Component ... · Bidirected Cut and Directed...
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Integrality Gaps ofBidirected Cut and Directed Component
Relaxations for Steiner Trees
Andreas Emil Feldmann Jochen Konemann Laura Sanita
University of WaterlooCombinatorics & Optimization
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The Steiner Tree Problem
Terminals
Steiner vertices
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The Steiner Tree Problem
Terminals
Steiner vertices
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Bidirected Cut Relaxation (BCR)
Terminals
Steiner vertices
[Edmonds 1967]
I add root
I bi-direct edges
I integral flow =Steiner tree
Relaxation:
I find capacities xa forflow demand 1 to root
I cost:∑
a∈A caxa
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Bidirected Cut Relaxation (BCR)
Terminals
Steiner vertices
root
[Edmonds 1967]
I add root
I bi-direct edges
I integral flow =Steiner tree
Relaxation:
I find capacities xa forflow demand 1 to root
I cost:∑
a∈A caxa
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Bidirected Cut Relaxation (BCR)
Terminals
Steiner vertices
root
[Edmonds 1967]
I add root
I bi-direct edges
I integral flow =Steiner tree
Relaxation:
I find capacities xa forflow demand 1 to root
I cost:∑
a∈A caxa
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Bidirected Cut Relaxation (BCR)
Terminals
Steiner vertices
root
[Edmonds 1967]
I add root
I bi-direct edges
I integral flow =Steiner tree
Relaxation:
I find capacities xa forflow demand 1 to root
I cost:∑
a∈A caxa
![Page 8: Integrality Gaps of Bidirected Cut and Directed Component ... · Bidirected Cut and Directed Component Relaxations for Steiner Trees Andreas Emil Feldmann Jochen K onemann Laura Sanit](https://reader033.fdocuments.in/reader033/viewer/2022041704/5e43da20b5227f59f061c39b/html5/thumbnails/8.jpg)
Directed Component Relaxation (DCR)
Terminals
Steiner vertices
[Polzin, Daneshmand 2003]
I add root
I integral flow throughfull components
= Steiner tree
Relaxation:
I find capacities xK offull components forflow demand 1 to root
I cost:∑
K∈K cKxK
(cK =∑
e∈K ce)
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Directed Component Relaxation (DCR)
Terminals
Steiner vertices
root
[Polzin, Daneshmand 2003]
I add root
I integral flow throughfull components
= Steiner tree
Relaxation:
I find capacities xK offull components forflow demand 1 to root
I cost:∑
K∈K cKxK
(cK =∑
e∈K ce)
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Directed Component Relaxation (DCR)
Terminals
Steiner vertices
root
[Polzin, Daneshmand 2003]
I add root
I integral flow throughfull components
= Steiner tree
Relaxation:
I find capacities xK offull components forflow demand 1 to root
I cost:∑
K∈K cKxK
(cK =∑
e∈K ce)
![Page 11: Integrality Gaps of Bidirected Cut and Directed Component ... · Bidirected Cut and Directed Component Relaxations for Steiner Trees Andreas Emil Feldmann Jochen K onemann Laura Sanit](https://reader033.fdocuments.in/reader033/viewer/2022041704/5e43da20b5227f59f061c39b/html5/thumbnails/11.jpg)
BCR v.s. DCR Gaps
DCR BCRINT INT
DCR gap ≤ ln(4) ≈ 1.39[Goemans et al. 2012]
BCR gap ≤ 2
DCR gap ≤ BCR gap
DCR gap?= BCR gap
NP-hard
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BCR v.s. DCR Gaps
DCR BCR
compositionmap
INT INT
DCR gap ≤ ln(4) ≈ 1.39[Goemans et al. 2012]
BCR gap ≤ 2
DCR gap ≤ BCR gap
DCR gap?= BCR gap
NP-hard
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BCR v.s. DCR Gaps
DCR BCR
compositionmap
INT INT
DCR gap ≤ ln(4) ≈ 1.39[Goemans et al. 2012]
BCR gap ≤ 2
DCR gap ≤ BCR gap
DCR gap?= BCR gap
NP-hard
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Quasi-Bipartite Graphs
Non-constructive:
Constructive:
DCR opt =
dual DCR opt ≤ dual BCR opt
= BCR opt
BCR opt ≥ DCR opt
[Chakrabarty et al. 2011]
[Goemans et al. 2012]
[Fung et al. 2012]
DCR gap = BCR gap
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Quasi-Bipartite Graphs
map
DCR BCR
dual DCR dual BCR
Non-constructive:
Constructive:
DCR opt =
dual DCR opt ≤ dual BCR opt
= BCR opt
BCR opt ≥ DCR opt
[Chakrabarty et al. 2011]
[Goemans et al. 2012]
[Fung et al. 2012]
DCR gap = BCR gap
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Quasi-Bipartite Graphs
decomposition
DCR BCR
dual DCR dual BCR
Non-constructive: Constructive:
DCR opt =
dual DCR opt ≤ dual BCR opt
= BCR opt
BCR opt ≥ DCR opt
[Chakrabarty et al. 2011] [Goemans et al. 2012]
[Fung et al. 2012]
DCR gap = BCR gap
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2-Quasi-Bipartite Graphs
Non-constructive:
DCR opt =
dual DCR opt ≤ dual BCR opt
= BCR opt
DCR gap = BCR gap
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2-Quasi-Bipartite Graphs
map
DCR BCR
dual DCR dual BCR
Non-constructive:
DCR opt =
dual DCR opt ≤ dual BCR opt
= BCR opt
DCR gap = BCR gap
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2-Quasi-Bipartite Graphs
decomposition
DCR BCR
dual DCR dual BCR
Constructive?
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2-Quasi-Bipartite Graphs
BCRdecomposableto DCR 2
2
2 3
3
1
1
3
3
1
1
Constructive?
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Further Generalizations
DCR gap v.s. BCR gap?
DCR opt
BCR opt=
12
11
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Further Generalizations
DCR gap v.s. BCR gap? DCR opt
BCR opt=
12
11
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Open Problems
I Constructive proof for 2-quasi-bipartite graphs?
I Equal gaps for k-quasi-bipartite graphs?(Conjecture: yes.)
I Gap difference for more general graphs?