Integrality Gaps ofBidirected Cut and Directed Component

Relaxations for Steiner Trees

Andreas Emil Feldmann Jochen Konemann Laura Sanita

University of WaterlooCombinatorics & Optimization

The Steiner Tree Problem

Terminals

Steiner vertices

The Steiner Tree Problem

Terminals

Steiner vertices

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

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

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

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

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)

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)

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)

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

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

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

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

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

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

2-Quasi-Bipartite Graphs

Non-constructive:

DCR opt =

dual DCR opt ≤ dual BCR opt

= BCR opt

DCR gap = BCR gap

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

2-Quasi-Bipartite Graphs

decomposition

DCR BCR

dual DCR dual BCR

Constructive?

2-Quasi-Bipartite Graphs

BCRdecomposableto DCR 2

2

2 3

3

1

1

3

3

1

1

Constructive?

Further Generalizations

DCR gap v.s. BCR gap?

DCR opt

BCR opt=

12

11

Further Generalizations

DCR gap v.s. BCR gap? DCR opt

BCR opt=

12

11

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?