Node Multiway Cut and Subset Feedback Vertex Set on Graphs of - PowerPoint PPT Presentation

Node Multiway Cut and Subset Feedback Vertex Set on Graphs of Bounded Mim-width Benjamin Bergougnoux ⋆ , Charis Papadopoulos † and Jan Arne Telle ⋆ . WG 2020, June 26, 2020 ⋆ University of Bergen, Norway † University of Ioannina, Greece

The parameters

Treewidth ◮ Unbounded in any dense graph class. ◮ Tractability results on dense graph classes can be explained with rank-width and mim-width.

Divide a graph Recursively decompose your graph... G

Divide a graph Recursively decompose your graph... into simple cuts. → We describe the simplicity of a cut with a function f: cut → Q . Different notions of simplicity = different width measures.

Divide a graph Width of a decomposition D := max f ( cut ) among the cuts of D . G

Divide a graph Width of a graph G := min width of a decomposition of G . G G G G

Rank-width Defined from the function rw ( A, B ) := the rank of adjacency matrix between A and B over GF (2) . B   1 0 1 rw 0 1 1  =2  1 1 0 A

Maximum Induced Matching width (mim-width) Defined from the function mim ( A, B ) := the size of a maximum induced matching in the bipartite graph between A and B . B A

Hierarchy Meta-algorithmic Modeling Power applications Interval, permutation, ( σ, ρ )-Dominating mim-width k -polygon, Dilworth-k, Set problems convex graphs,... MSO 1 clique-width Cographs, rank-width hereditary distance MSO 2 Tree, tree-width partial k -tree

Computation complexity ◮ NP-hard for all these widths measures. ◮ Efficient algorithms for tree-width and rank-width. → Running time: 2 O ( k ) · n O (1) . ◮ Tough open question for mim-width.

The problem

Feedback Vertex Set Feedback Vertex Set Find a set of vertices X of minimum weight hitting all cycles ( G − X is a forest). X

Subset Feedback Vertex Set Subset Feedback Vertex Set Find a set of vertices X of minimum weight hitting all cycles going through a prescribed set of vertices S . S X

Subset Feedback Vertex Set Subset Feedback Vertex Set Find a set of vertices X of minimum weight hitting all cycles going through a prescribed set of vertices S . S

Subset FVS is harder Corneil and Fonlupt 1988 FVS is solvable in polynomial time on split graphs. Fomin et al. 2014 Subset FVS is NP-hard on split graphs. Bodalender, Cygan, Kratsch and Nederlof 2015 FVS is solvable in time 2 O ( tw ) · n . B., Bonnet, Brettell and Kwon [ArxiV] Subset FVS can not be solved in time 2 o ( tw log tw ) · n O (1) unless ETH fails.

Recent results Papadopoulos and Tzimas 2019 Subset FVS is solvable in polynomial time on interval graphs and permutation graphs. Jaffke, Kwon and Telle 2019 | B. and Kanté 2019 FVS is solvable in time n O ( mim ) given a decomposition. Our result Subset FVS is solvable in time n O ( mim 2 ) given a decomposition.

Algorithmic consequences Corollary Subset FVS is solvable in polynomial time on interval, permutation, bi-interval graphs, circular arc and circular permutation graphs, convex graphs, k -polygon, Dilworth- k and co- k -degenerate graphs for fixed k . Theorem Subset FVS is solvable in time 2 O ( rw 3 ) · n O (1) . Same results holds for Node Multiway Cut.

Our Approach

Neighbor equivalence Definition X, Y ⊆ A are 1 -neighbor equivalent over A if N ( X ) ∩ B = N ( Y ) ∩ B. nec 1 ( A ) := number of equivalence classes over A . B A X Y Lemma For every A ⊆ V ( G ) , have nec 2 ( A ) ≤ n O ( mim ) .

Reduce Lemma It is sufficient to keep for each cut ( A, B ) ( nec 2 ( A ) + nec 2 ( B )) O ( mim ) · mim O ( mim ) ≤ n O ( mim 2 ) partial solutions. |D| ≤ n O ( mim 2 ) Set of partial D ⊆ C solutions Reduce small and equivalent C ⊆ 2 A to C ∀ Y ⊆ B , best ( C , Y ) = best ( D , Y ) Corollary Subset FVS is solvable in time n O ( mim 2 ) given a decomposition.

S-forest Minimum Subset FVS Complement: S -forest

S-forest Minimum Subset FVS Complement ↓ : forest

S-forest − → forest Lemma We can contract the components of G [ X ] − S and unions of the components of G [ Y ] − S such that the resulting graph is a forest. Y B A X

S-forest − → forest Lemma We can contract the components of G [ X ] − S and unions of the components of G [ Y ] − S such that the resulting graph is a forest. Y ↓ B A X ↓

S-forest − → forest Lemma We can contract the components of G [ X ] − S and unions of the components of G [ Y ] − S such that the resulting graph is a forest. Y ↓ B A X ↓

Small vertex cover Lemma [B. and Kanté 2019] After contractions, every induced forest between A and B have a vertex cover of size 4 mim whose vertices have different neighborhoods. Y ↓ B A X ↓

Indices Definition An index i is a collection of at most 4 mim + 1 representatives for the 2-neighbor equivalence over A and B . Number of indices ≤ ( nec 2 ( A ) + nec 2 ( B )) O ( mim ) ≤ n O ( mim 2 ) . Y ↓ B A X ↓ X \ V C i

Index association We associate each index with some partial solutions and some subsets of B . B A X \ V C X ↓ i

Index association We associate each index with some partial solutions and some subsets of B . Y \ V C Y ↓ B A i

Index association Lemma For every solution F , there exists an index i such that F ∩ A and F ∩ B are associated with i . F ∩ B ↓ B A F ∩ A ↓ i

Index association Definition Two partial solutions X, W associated with i are i -equivalent if they connect the vertices of the vertex cover in the same way. Number of i -equivalence classes ≤ mim O ( mim ) Lemma If X and W are i -equivalent, then for every Y ⊆ B associated with i , G [ X ∪ Y ] is a solution iff G [ W ∪ Y ] is also a solution.

Result Lemma It is sufficient to keep, for every index i , a partial solutions of maximum weight for each i -equivalence class. The number of partial solutions we keep is at most ( nec 2 ( A ) + nec 2 ( B )) O ( mim 2 ) mim O ( mim ) ≤ n O ( mim 2 ) Theorem Subset FVS is solvable in time n O ( mim 2 ) given a decomposition.

Algorithmic consequences We have nec 2 ( A ) ≤ 2 O ( rw 2 ) and mim ≤ rw Lemma It is sufficient to keep, for every index i , a partial solutions of maximum weight for each i -equivalence class. The number of partial solutions we keep is at most ( nec 2 ( A ) + nec 2 ( B )) O ( mim 2 ) mim O ( mim ) ≤ 2 O ( rw 3 ) Theorem Subset FVS is solvable in time 2 O ( rw 3 ) · n O (1) .

Node Multiway Cut T X Theorem Node Multiway Cut is solvable in time 2 O ( rw 3 ) and n O ( mim 2 ) given a decomposition.

Node Multiway Cut T Theorem Node Multiway Cut is solvable in time 2 O ( rw 3 ) and n O ( mim 2 ) given a decomposition.

Node Multiway Cut S T X Theorem Node Multiway Cut is solvable in time 2 O ( rw 3 ) and n O ( mim 2 ) given a decomposition.

Neighbor equivalence relations These relations have been used for many problems: ◮ ( σ, ρ ) -Dominating Set problems [Bui-Xuan, Telle and Vatshelle 2013] ◮ Their connected and acyclic variants and Max Cut [B. and Kanté 2019] ◮ Subset FVS and Node Multiway Cut [B., Papadapoulos and Telle] for many parameters: tree-width, clique-width, rank-width, Q -rank-width, mim-width

Open questions ◮ Can we characterize which problems are in XP parameterized by the mim-width of a given decomposition? B. and Kanté 2019 The connected and acyclic variants of ( σ, ρ ) -Dominating Set problems are solvable in time n O ( mim ) given a decomposition. ◮ Can we approximate/compute the mim-width of a graph in XP-time? ◮ Can we recognize the graphs of mim-width one?

Casual dynamic programming G

Combine and Reduce S Reduction { X 1 ∪ X 2 | X 1 ∈ S 1 , X 2 ∈ S 2 } S 1 S 2 Combination

Reduce: tough and key step In general If it is enough to keep N partial solutions at each step, then we can solve the problem in time N O (1) · n O (1) .