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 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0   0 0 1 1 0 0 0 0 0 0     0 0 0 1 1 0 0 0 0 0     1 0 0 0 1 0 0 0 0 0     1 0 0 0 0 1 0 0 0 0    0 1 0 0 0 0 1 0 0 0    0 0 1 0 0 0 0 1 0 0     0 0 0 1 0 0 0 0 1 0     0 0 0 0 1 0 0 0 0 1     0 0 0 0 0 1 1 0 0 0     0 0 0 0 0 0 1 1 0 0     0 0 0 0 0 0 0 1 1 0   0 0 0 0 0 0 0 0 1 1   0 0 0 0 0 1 0 0 0 1 What is the maximum sub-determinant of M ?

Puzzle: G graph, M = M ( G ) vertex-edge incidence matrix of G   1 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0   0 0 1 1 0 0 0 0 0 0     0 0 0 1 1 0 0 0 0 0     1 0 0 0 1 0 0 0 0 0     1 0 0 0 0 1 0 0 0 0    0 1 0 0 0 0 1 0 0 0    0 0 1 0 0 0 0 1 0 0     0 0 0 1 0 0 0 0 1 0     0 0 0 0 1 0 0 0 0 1     0 0 0 0 0 1 1 0 0 0     0 0 0 0 0 0 1 1 0 0     0 0 0 0 0 0 0 1 1 0   0 0 0 0 0 0 0 0 1 1   0 0 0 0 0 1 0 0 0 1 What is the maximum sub-determinant of M ?

Puzzle: G graph, M = M ( G ) vertex-edge incidence matrix of G   1 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0   0 0 1 1 0 0 0 0 0 0     0 0 0 1 1 0 0 0 0 0     1 0 0 0 1 0 0 0 0 0     1 0 0 0 0 1 0 0 0 0    0 1 0 0 0 0 1 0 0 0    0 0 1 0 0 0 0 1 0 0     0 0 0 1 0 0 0 0 1 0     0 0 0 0 1 0 0 0 0 1     0 0 0 0 0 1 1 0 0 0     0 0 0 0 0 0 1 1 0 0     0 0 0 0 0 0 0 1 1 0   0 0 0 0 0 0 0 0 1 1   0 0 0 0 0 1 0 0 0 1 What is the maximum sub-determinant of M ?

Puzzle: G graph, M = M ( G ) vertex-edge incidence matrix of G   1 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0   0 0 1 1 0 0 0 0 0 0     0 0 0 1 1 0 0 0 0 0     1 0 0 0 1 0 0 0 0 0     1 0 0 0 0 1 0 0 0 0    0 1 0 0 0 0 1 0 0 0    0 0 1 0 0 0 0 1 0 0     0 0 0 1 0 0 0 0 1 0     0 0 0 0 1 0 0 0 0 1     0 0 0 0 0 1 1 0 0 0     0 0 0 0 0 0 1 1 0 0     0 0 0 0 0 0 0 1 1 0   0 0 0 0 0 0 0 0 1 1   0 0 0 0 0 1 0 0 0 1 What is the maximum sub-determinant of M ? Answer: ∆ = 2 · 2

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

Lemma (Folklore) If G is a graph and ∆ is the maximum sub-determinant of M ( G ) , then ∆ = 2 OCP( G ) where OCP( G ) is the odd cycle packing number of G = ⇒ OCP( G ) should be a “good” complexity measure for the IP � max c v x v v ∈ V ( G ) s . t . Mx � 1 x � 0 x ∈ Z V ( G )

Lemma (Folklore) If G is a graph and ∆ is the maximum sub-determinant of M ( G ) , then ∆ = 2 OCP( G ) where OCP( G ) is the odd cycle packing number of G = ⇒ OCP( G ) should be a “good” complexity measure for the IP � max c v x v v ∈ V ( G ) s . t . Mx � 1 x � 0 x ∈ Z V ( G ) which is a generic instance of the maximum weight stable set problem

What can we expect? (Widely open) Conjecture For every fixed k ∈ Z � 0 , the maximum weight stable set problem can be solved in polynomial time on graphs G with OCP( G ) � k

What can we expect? (Widely open) Conjecture For every fixed k ∈ Z � 0 , 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 ∈ Z � 1 , IPs with maximum sub-determinant ∆ � K can be solved in polynomial time

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

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

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 ∈ Z � 0 , the maximum weight stable set problem on graphs G with OCP( G ) � k has a PTAS

Escher wall graph:

Escher wall graph:

Escher wall graph:

Escher wall graph:

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 − { e 1 , e 2 , e 3 } is bipartite for some 3 -cycle { e 1 , e 2 , e 3 } ⊆ E ( G ) , (iv) G has an even face embedding in the projective plane

What should we do? Efficient algorithms Nice polyhedra Min-Max relations

What should we do? Efficient algorithms Nice polyhedra Min-Max relations Stable set polytope : STAB( G ) := conv { χ S ∈ { 0 , 1 } V ( G ) | S ⊆ V ( G ) stable set of G } = conv { x ∈ Z V ( G ) | Mx � 1 , x � 0 }

What should we do? Efficient algorithms Nice polyhedra Min-Max relations Stable set polytope : STAB( G ) := conv { χ S ∈ { 0 , 1 } V ( G ) | S ⊆ V ( G ) stable set of G } = conv { x ∈ Z V ( G ) | Mx � 1 , x � 0 } Our goal Find a polynomial-size extended formulation of STAB( G ) for all graphs G with OCP( G ) = 1

Our strategy Construct an extended formulation of 1 P ( G ) := conv { x ∈ Z V ( G ) | Mx � 1 }

Our strategy Construct an extended formulation of 1 P ( G ) := conv { x ∈ Z V ( G ) | Mx � 1 } Use fact that STAB( G ) = P ( G ) ∩ [0 , 1] V ( G ) for all graphs G 2

Our strategy Construct an extended formulation of 1 P ( G ) := conv { x ∈ Z V ( G ) | Mx � 1 } Use fact that STAB( G ) = P ( G ) ∩ [0 , 1] V ( G ) for all graphs G 2 Work in slack space with variables y := 1 − Mx ∈ R E ( G ) 3

Our strategy Construct an extended formulation of 1 P ( G ) := conv { x ∈ Z V ( G ) | Mx � 1 } Use fact that STAB( G ) = P ( G ) ∩ [0 , 1] V ( G ) for all graphs G 2 Work in slack space with variables y := 1 − Mx ∈ R E ( G ) 3 Focus (first) on case where G has an even-face embedding in P 4

it i The slack map Assume: G is 2 -connected and non-bipartite

it i The slack map Assume: G is 2 -connected and non-bipartite For x ∈ R V ( G ) , let y := 1 − Mx ∈ R E ( G )

it i The slack map Assume: G is 2 -connected and non-bipartite For x ∈ R V ( G ) , let y := 1 − Mx ∈ R E ( G ) ⇐ ⇒ y vw = 1 − x v − x w for all edges vw ∈ E ( G )

it i The slack map Assume: G is 2 -connected and non-bipartite For x ∈ R V ( G ) , let y := 1 − Mx ∈ R E ( G ) ⇐ ⇒ y vw = 1 − x v − x w for all edges vw ∈ E ( G ) Slack map σ : R V ( G ) → R E ( G ) : x �→ y = σ ( x ) := 1 − Mx

it The slack map Assume: G is 2 -connected and non-bipartite For x ∈ R V ( G ) , let y := 1 − Mx ∈ R E ( G ) ⇐ ⇒ y vw = 1 − x v − x w for all edges vw ∈ E ( G ) Slack map σ : R V ( G ) → R E ( G ) : x �→ y = σ ( x ) := 1 − Mx For every even cycle C = ( e 1 , e 2 , . . . , e 2 k ) , have i 2 k 2 k � � ( − 1) i − 1 y e i = ( − 1) i − 1 (1 − x v i − x v i +1 ) i =1 i =1 = x v 2 k +1 − x v 1 = 0

Lemma The image σ ( R V ( G ) ) of the slack map is the linear subspace of R E ( G ) defined by 2 k � ( − 1) i − 1 y e i = 0 ∀ even cycles C = ( e 1 , e 2 , . . . , e 2 k ) i =1

If x ∈ Z V ( G ) , for every odd cycle C = ( e 1 , e 2 , . . . , e 2 k +1 ) : 2 k +1 2 k +1 � � y e i = (1 − x v i − x v i +1 ) i =1 i =1 2 k +1 � = | C | − 2 x v i i =1 ≡ 1 (mod 2)

If x ∈ Z V ( G ) , for every odd cycle C = ( e 1 , e 2 , . . . , e 2 k +1 ) : 2 k +1 2 k +1 � � y e i = (1 − x v i − x v i +1 ) i =1 i =1 2 k +1 � = | C | − 2 x v i i =1 ≡ 1 (mod 2) Lemma For any fixed odd cycle C 0 in G ,     y ∈ Z V ( G ) |  � σ ( Z V ( G ) ) = σ ( R V ( G ) ) ∩ y e is odd  e ∈ E ( C 0 )

Theorem (Slack representation) For any fixed odd cycle C 0 in G ,     y ∈ Z V ( G ) |  �  ∩ R E ( G ) σ ( Z V ( G ) ∩ P ( G )) = σ ( R V ( G ) ) ∩ y e is odd + e ∈ E ( C 0 )