Stable sets in graphs with no two disjoint odd cycles Michele - - PowerPoint PPT Presentation

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Stable sets in graphs with no two disjoint odd cycles Michele - - PowerPoint PPT Presentation

Stable sets in graphs with no two disjoint odd cycles Michele Conforti S. F . Tony Huyhn Stefan Weltge 23rd Aussois Combinatorial Optimization Workshop, 8th Jan. 2019 Puzzle: G graph, M = M ( G ) vertex-edge incidence matrix of G 1


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Stable sets in graphs with no two disjoint

  • dd cycles

Michele Conforti

  • S. F

. Tony Huyhn Stefan Weltge

23rd Aussois Combinatorial Optimization Workshop, 8th Jan. 2019

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Puzzle: G graph, M = M(G) vertex-edge incidence matrix of G                       

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

                       What is the maximum sub-determinant of M?

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Puzzle: G graph, M = M(G) vertex-edge incidence matrix of G                       

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

                       What is the maximum sub-determinant of M?

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Puzzle: G graph, M = M(G) vertex-edge incidence matrix of G                       

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

                       What is the maximum sub-determinant of M?

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Puzzle: G graph, M = M(G) vertex-edge incidence matrix of G                       

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

                       What is the maximum sub-determinant of M? Answer: ∆ = 2 · 2

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Lemma (Folklore) If G is a graph and ∆ is the maximum sub-determinant of M(G), then ∆ = 2OCP(G) where OCP(G) is the odd cycle packing number of G

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Lemma (Folklore) If G is a graph and ∆ is the maximum sub-determinant of M(G), then ∆ = 2OCP(G) where OCP(G) is the odd cycle packing number of G = ⇒ OCP(G) should be a “good” complexity measure for the IP max

  • v∈V (G)

cvxv s.t. Mx 1 x 0 x ∈ ZV (G)

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Lemma (Folklore) If G is a graph and ∆ is the maximum sub-determinant of M(G), then ∆ = 2OCP(G) where OCP(G) is the odd cycle packing number of G = ⇒ OCP(G) should be a “good” complexity measure for the IP max

  • v∈V (G)

cvxv s.t. Mx 1 x 0 x ∈ ZV (G) which is a generic instance of the maximum weight stable set problem

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What can we expect?

(Widely open) Conjecture For every fixed k ∈ Z0, the maximum weight stable set problem can be solved in polynomial time on graphs G with OCP(G) k

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What can we expect?

(Widely open) Conjecture For every fixed k ∈ Z0, the maximum weight stable set problem can be solved in polynomial time on graphs G with OCP(G) k Would follow from: (Even more crazy) Conjecture For every fixed K ∈ Z1, IPs with maximum sub-determinant ∆ K can be solved in polynomial time

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What is known?

Theorem (Artmann, Weismantel, Zenklusen STOC’17) Bimodular IPs can be solved in strongly polynomial time

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What is known?

Theorem (Artmann, Weismantel, Zenklusen STOC’17) Bimodular IPs can be solved in strongly polynomial time Corollary The maximum stable set problem can be solved in strongly polynomial time on graphs G with OCP(G) 1

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What is known?

Theorem (Artmann, Weismantel, Zenklusen STOC’17) Bimodular IPs can be solved in strongly polynomial time Corollary The maximum stable set problem can be solved in strongly polynomial time on graphs G with OCP(G) 1 Theorem (Bock, Faenza, Moldenhauer, Vargas, Jacinto’14) For every fixed k ∈ Z0, the maximum weight stable set problem on graphs G with OCP(G) k has a PTAS

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Escher wall graph:

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Escher wall graph:

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Escher wall graph:

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Escher wall graph:

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Theorem (Lov´ asz, cited in Seymour’95) Let G be an internally 4-connected graph. Then G has no two vertex-disjoint odd cycles if and only if G satisfies one of the following: (i) |G| 5, (ii) G − {x} is bipartite for some x ∈ V (G), (iii) G − {e1, e2, e3} is bipartite for some 3-cycle {e1, e2, e3} ⊆ E(G), (iv) G has an even face embedding in the projective plane

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What should we do?

Efficient algorithms Min-Max relations Nice polyhedra

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What should we do?

Efficient algorithms Min-Max relations Nice polyhedra

Stable set polytope: STAB(G) := conv{χS ∈ {0, 1}V (G) | S ⊆ V (G) stable set of G} = conv{x ∈ ZV (G) | Mx 1, x 0}

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What should we do?

Efficient algorithms Min-Max relations Nice polyhedra

Stable set polytope: STAB(G) := conv{χS ∈ {0, 1}V (G) | S ⊆ V (G) stable set of G} = conv{x ∈ ZV (G) | Mx 1, x 0} Our goal Find a polynomial-size extended formulation of STAB(G) for all graphs G with OCP(G) = 1

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Our strategy

1

Construct an extended formulation of P(G) := conv{x ∈ ZV (G) | Mx 1}

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Our strategy

1

Construct an extended formulation of P(G) := conv{x ∈ ZV (G) | Mx 1}

2

Use fact that STAB(G) = P(G) ∩ [0, 1]V (G) for all graphs G

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Our strategy

1

Construct an extended formulation of P(G) := conv{x ∈ ZV (G) | Mx 1}

2

Use fact that STAB(G) = P(G) ∩ [0, 1]V (G) for all graphs G

3

Work in slack space with variables y := 1 − Mx ∈ RE(G)

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Our strategy

1

Construct an extended formulation of P(G) := conv{x ∈ ZV (G) | Mx 1}

2

Use fact that STAB(G) = P(G) ∩ [0, 1]V (G) for all graphs G

3

Work in slack space with variables y := 1 − Mx ∈ RE(G)

4

Focus (first) on case where G has an even-face embedding in P

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The slack map

Assume: G is 2-connected and non-bipartite

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The slack map

Assume: G is 2-connected and non-bipartite For x ∈ RV (G), let y := 1 − Mx ∈ RE(G)

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The slack map

Assume: G is 2-connected and non-bipartite For x ∈ RV (G), let y := 1 − Mx ∈ RE(G) ⇐ ⇒ yvw = 1 − xv − xw for all edges vw ∈ E(G)

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The slack map

Assume: G is 2-connected and non-bipartite For x ∈ RV (G), let y := 1 − Mx ∈ RE(G) ⇐ ⇒ yvw = 1 − xv − xw for all edges vw ∈ E(G) Slack map σ : RV (G) → RE(G) : x → y = σ(x) := 1 − Mx

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The slack map

Assume: G is 2-connected and non-bipartite For x ∈ RV (G), let y := 1 − Mx ∈ RE(G) ⇐ ⇒ yvw = 1 − xv − xw for all edges vw ∈ E(G) Slack map σ : RV (G) → RE(G) : x → y = σ(x) := 1 − Mx For every even cycle C = (e1, e2, . . . , e2k), have

2k

  • i=1

(−1)i−1yei =

2k

  • i=1

(−1)i−1(1 − xvi − xvi+1) = xv2k+1 − xv1 = 0

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Lemma The image σ(RV (G)) of the slack map is the linear subspace of RE(G) defined by

2k

  • i=1

(−1)i−1yei = 0 ∀even cycles C = (e1, e2, . . . , e2k)

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If x ∈ ZV (G), for every odd cycle C = (e1, e2, . . . , e2k+1):

2k+1

  • i=1

yei =

2k+1

  • i=1

(1 − xvi − xvi+1) = |C| − 2

2k+1

  • i=1

xvi ≡ 1 (mod 2)

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If x ∈ ZV (G), for every odd cycle C = (e1, e2, . . . , e2k+1):

2k+1

  • i=1

yei =

2k+1

  • i=1

(1 − xvi − xvi+1) = |C| − 2

2k+1

  • i=1

xvi ≡ 1 (mod 2) Lemma For any fixed odd cycle C0 in G, σ(ZV (G)) = σ(RV (G)) ∩   y ∈ ZV (G) |

  • e∈E(C0)

ye is odd   

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Theorem (Slack representation) For any fixed odd cycle C0 in G, σ(ZV (G)∩P(G)) = σ(RV (G))∩   y ∈ ZV (G) |

  • e∈E(C0)

ye is odd   ∩RE(G)

+

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Assume: G is embedded in the projective plane with all faces even

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Assume: G is embedded in the projective plane with all faces even

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Assume: G is embedded in the projective plane with all faces even

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Fact: The dual graph G∗ has a canonical orientation

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Regard slack map as map σ : RV (G) → RE(G∗)

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Regard slack map as map σ : RV (G) → RE(G∗) Equations definining σ(RV (G)):

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Regard slack map as map σ : RV (G) → RE(G∗) Equations definining σ(RV (G)):

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= ⇒ σ(RV (G)) ⊆ {y ∈ RE(G∗) | y circulation}

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Regard slack map as map σ : RV (G) → RE(G∗) Equations definining σ(RV (G)):

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= ⇒ σ(RV (G)) ⊆ {y ∈ RE(G∗) | y circulation} Euler’s formula: |E(G)| − |V (G)| = |V (G∗)| − 1

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Regard slack map as map σ : RV (G) → RE(G∗) Equations definining σ(RV (G)):

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= ⇒ σ(RV (G)) ⊆ {y ∈ RE(G∗) | y circulation} Euler’s formula: |E(G)| − |V (G)| = |V (G∗)| − 1 = ⇒ σ(RV (G)) = {y ∈ RE(G∗) | y circulation}

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Regard slack map as map σ : RV (G) → RE(G∗) Equations definining σ(RV (G)):

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= ⇒ σ(RV (G)) ⊆ {y ∈ RE(G∗) | y circulation} Euler’s formula: |E(G)| − |V (G)| = |V (G∗)| − 1 = ⇒ σ(RV (G)) = {y ∈ RE(G∗) | y circulation} Lemma Let C0 be any fixed odd cycle in G. The points in σ(ZV (G) ∩ P(G)) are exactly the integer, non-negative circulations y in G∗ such that

  • e∈E(C0) ye∗ is odd
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To get an extended formulation of P(G) we use disjunctive programming (in order to guess a dual vertex) the standard “double cover” construction

:* :

: :* :

.

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To get an extended formulation of P(G) we use disjunctive programming (in order to guess a dual vertex) the standard “double cover” construction

:* :

: :* :

.

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To get an extended formulation of P(G) we use disjunctive programming (in order to guess a dual vertex) the standard “double cover” construction

:* :

: :* :

.

Theorem If G is a 2-connected, non-bipartite graph embedded in the projective plane P with all faces even then STAB(G) has a O(n2)-size extended formulation

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Higher genus surfaces

Fact: if G is embedded on a fixed non-orientable surface S with cross-cap number k in such a way that every odd cycle is 1-sided (that is, admits a neighborhood that is a M¨

  • bius band), then

OCP(G) k

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Higher genus surfaces

Fact: if G is embedded on a fixed non-orientable surface S with cross-cap number k in such a way that every odd cycle is 1-sided (that is, admits a neighborhood that is a M¨

  • bius band), then

OCP(G) k Theorem If G can be embedded in the Klein bottle K with all odd cycles 1-sided then STAB(G) has a polynomial-size extended formulation

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Higher genus surfaces

Fact: if G is embedded on a fixed non-orientable surface S with cross-cap number k in such a way that every odd cycle is 1-sided (that is, admits a neighborhood that is a M¨

  • bius band), then

OCP(G) k Theorem If G can be embedded in the Klein bottle K with all odd cycles 1-sided then STAB(G) has a polynomial-size extended formulation Conjecture If G can be embedded in any fixed surface S with all odd cycles 1-sided then STAB(G) has a polynomial-size extended formulation

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Thank you!