More applications of the d -neighbor equivalence: acyclic and connectivity constraints Benjamin Bergougnoux ⋆ , Mamadou Moustapha Kanté † ⋆ IRIF, CNRS, Université Paris Diderot, France † LIMOS, CNRS, Université Clermont Auvergne, France ESA, Munich, October 11, 2019
Width measures ▸ Tree-width is nice but unbounded in any dense graph class. ▸ Many NP-hard problems are tractable on some dense graph classes. → Explainable with clique-width, rank-width, mim width. 2 / 29
Divide a graph Recursively decompose your graph... G We recursively cut the vertex set in two 3 / 29
Divide a graph Recursively decompose your graph... into simple cuts. → We describe the simplicity of a cut with a function f: cut → N . Different notions of simplicity = different width measures. 3 / 29
Divide a graph Width of a decomposition D ∶ = max f ( cut ) among the cuts of D . G 3 / 29
Divide a graph Width of a graph G ∶ = min width of the decompositions of G . G G G G 3 / 29
Module-width Defined from the function mw ( A ) ∶= ∣{ N ( v ) ∩ A ∶ v ∈ A }∣ . A A Linearly equivalent to clique-width: [Rao 2006] For all graphs G , we have mw ( G ) ≤ cw ( G ) ≤ 2 mw ( G ) . 4 / 29
Rank-width Defined from the function rw ( A ) ∶= the rank of adjacency matrix between A and A over GF ( 2 ) . A 1 0 1 0 1 1 1 1 0 A 5 / 29
Maximum Induced Matching width (mim-width) Defined from the function mim ( A ) ∶= the size of a maximum induced matching in the bipartite graph between A and A . A A 6 / 29
Generality / Modeling power permutation, interval, mim-width k -polygon, ... cliques, Clique, hereditary hereditary clique-width rank-width distance distance Tree, partial k -tree trees, partial k -trees tree-width 7 / 29
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 questions for clique-width and mim-width. 8 / 29
Algorithmic applications mim-width clique-width MSO 1 rank-width MSO 2 tree-width 9 / 29
Intuition: conquer G 10 / 29
Intuition: conquer S Reduction S 1 S 2 Combination 11 / 29
Intuition: conquer 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 ) . 12 / 29
One algorithm to rule them all Theorem [B., Kanté 2019] We have a meta-algorithm for the connected and acyclic variants of ( σ,ρ ) -dominating set problems. ▸ Connected dominating set, ▸ Maximum induced tree, ▸ Connected vertex cover, ▸ Longest induced path, ▸ Node-weighted Steiner tree, ▸ Maximum induced linear forest, ▸ Feedback vertex set, ▸ Max. induced tree of ∆ ≤ 42 ,... 13 / 29
One algorithm to rule them all Theorem [B., Kanté 2019] We have a meta-algorithm for the connected and acyclic variants of ( σ,ρ ) -dominating set problems. Corollary [B., Kanté 2019] These problems are solvable in time: 2 O ( tw ) ⋅ n O ( 1 ) tree-width 2 O ( cw ) ⋅ n O ( 1 ) clique-width 2 O ( rw 2 ) ⋅ n O ( 1 ) rank-width n O ( mim ) mim-width � ⇒ Polytime in interval graphs, permutations graphs, k -trapezoid,... 13 / 29
Key: d -neighbor equivalence nec d ( A ) : # of equivalence classes of the d -neighbor equivalence over A . Theorem [Bui-Xuan, Telle, Vatshelle 2013] It is enough to keep nec d ( A ) ⋅ nec d ( A ) partial solutions at each cut ( A,A ) , for any ( σ,ρ ) -Dominating Set problem. Lemma [Vatshelle 2012] nec d ( A ) is upper bounded by: 2 d ⋅ tw ⋅ n O ( 1 ) tree-width 2 d ⋅ cw ⋅ n O ( 1 ) clique-width 2 d ⋅ rw 2 ⋅ n O ( 1 ) rank-width n d ⋅ mim mim-width 14 / 29
Key: d -neighbor equivalence nec d ( A ) : # of equivalence classes of the d -neighbor equivalence over A . Theorem [Bui-Xuan, Telle, Vatshelle 2013] It is enough to keep nec d ( A ) ⋅ nec d ( A ) partial solutions at each cut ( A,A ) , for any ( σ,ρ ) -Dominating Set problem. Corollary [Bui-Xuan, Telle, Vatshelle 2013] We can solve these problems in time: 2 O ( tw ) ⋅ n O ( 1 ) tree-width 2 O ( cw ) ⋅ n O ( 1 ) clique-width 2 O ( rw 2 ) ⋅ n O ( 1 ) rank-width n O ( mim ) mim-width 14 / 29
Our result Theorem [B., Kanté 2019] It is enough to keep nec d ( A ) ⋅ nec d ( A ) ⋅ nec 1 ( A ) 2 partial solutions at each cut ( A,A ) , for any connected variant of a ( σ,ρ ) -Dominating Set problem. Corollary [B., Kanté 2019] We can solve these problems in time: 2 O ( tw ) ⋅ n O ( 1 ) tree-width 2 O ( cw ) ⋅ n O ( 1 ) clique-width 2 O ( rw 2 ) ⋅ n O ( 1 ) rank-width n O ( mim ) mim-width 15 / 29
Connected Dominating set and 1-neighbor equivalence 16 / 29
Connected Dominating set and 1-neighbor width Connected dominating set Finding a vertex set D of minimum weight which dominates all the vertices and which induces a connected graph. 17 / 29
1-neighbor equivalence relation X,Y ⊆ A are 1 -neighbor equivalent in A if N ( X ) ∩ A = N ( Y ) ∩ A . A A X Y nec 1 ( A ) ∶ = # of equivalence classes over A . Theorem [B. and Kanté 2018] It is enough to keep nec 1 ( A ) 3 nec 1 ( A ) partial solutions for each cut ( A,A ) to solve Connected dominating set. 18 / 29
Dealing with domination A set of partial solutions for each equivalence class R ′ of the 1- neighbor equivalence over A . A Y A 19 / 29
Dealing with domination The sets Y ∈ R ′ have the same neighborhood in A . A Y A 19 / 29
Dealing with domination The sets Y ∈ R ′ have the same neighborhood in A . A Y ′ A 19 / 29
Dealing with domination Partial solutions associated with R ′ : → X ⊆ A such that X ∪ Y dominates A , for (all) Y ∈ R ′ A Y X A 19 / 29
Dealing with domination Partial solutions associated with R ′ : → X ⊆ A such that X ∪ Y dominates A , for (all) Y ∈ R ′ A Y X ′ A 19 / 29
Dealing with domination At the root of the decomposition, the cut is ( V ( G ) , ∅ ) : → A partial solution is a dominating set. ∅ V ( G ) X 19 / 29
Dealing with domination Two partial solutions associated with R ′ are equivalent for the dom- ination if they are 1-neighbor equivalent! A Y X A 19 / 29
Dealing with domination Two partial solutions associated with R ′ are equivalent for the dom- ination if they are 1-neighbor equivalent! A Y X ′ A 19 / 29
Lemma [Bui-Xuan, Telle and Vatshelle, 2013] For Dominating set, it is enough to keep nec 1 ( A ) ⋅ nec 1 ( A ) partial solutions at each step. One partial solution: ▸ for each 1-neighbor equivalence class R ′ of A , and ▸ for each 1-neighbor equivalence class R of A 20 / 29
Dealing with connectivity ▸ We need an equivalence relation between sets of partial solutions. 21 / 29
Dealing with connectivity ▸ We need an equivalence relation between sets of partial solutions. ▸ For all Y ∈ R ′ and for all sets of partial solutions S , we define: best (S ,Y ) ∶ = min { weight ( X ) ∶ X ∈ S and G [ X ∪ Y ] is connected } . R ′ -representativity We say that S ⋆ R ′ -represents S if, for all Y ∈ R ′ , we have best (S ,Y ) = best (S ⋆ ,Y ) . 21 / 29
Representative set ▸ At the root ( V ( G ) , ∅ ) : an { ∅ } -representative set of the set of all partial solutions must contain an optimal solution. → best (S , ∅ ) ∶ = min { weigth ( X ) ∶ X ∈ S and G [ X ] is connected } . 22 / 29
Representative set ▸ At the root ( V ( G ) , ∅ ) : an { ∅ } -representative set of the set of all partial solutions must contain an optimal solution. → best (S , ∅ ) ∶ = min { weigth ( X ) ∶ X ∈ S and G [ X ] is connected } . Theorem [B., Kanté 2019] There exists a function reduce: ▸ Input: a set of partial solutions S ⊆ 2 A . ▸ Output: S ⋆ ⊆ S such that ∣ S ⋆ ∣ ≤ nec 1 ( A ) 2 and S ⋆ R ′ -represents S . ▸ Running time: ∣ S ∣ ⋅ nec 1 ( A ) O ( 1 ) ⋅ n 2 . 22 / 29
Sketch of proof Inspiration [Bodlaender, Cygan, Kratsch, Nederlof 2013] Rank based approach: technique to obtained 2 O ( tw ) ⋅ n time algorithms for many connectivity problems. Let M be the (S ,R ′ ) -matrix: ⎧ ⎪ ⎪ 1 if G [ X ∪ Y ] is connected , M[ X,Y ] ∶ = ⎨ ⎪ ⎪ 0 otherwise . ⎩ ▸ In GF ( 2 ) : a basis of the row space of M of minimum weight R ′ -represents S . ▸ But M is too big to be computed. 23 / 29
Sketch of proof P : the set of pairs ( R ′ 1 ,R ′ 2 ) of 1-neighbor equivalence classes in A . R ′ P R ′ GF (2) C S P C S M = ▸ A basis of C is also a basis of M . ▸ ∣ P ∣ = nec 1 ( A ) 2 . ▸ C is computable in time ∣ S ∣ ⋅ ∣ P ∣ ⋅ n 2 . 24 / 29
Trick There is 2 | CC ( X ∪ Y ) |− 1 ways of dividing X ∪ Y such that v Y is on one side. v Y 25 / 29
Trick There is 2 | CC ( X ∪ Y ) |− 1 ways of dividing X ∪ Y such that v Y is on one side. v Y C · C [ X, Y ] = 2 | CC ( X ∪ Y ) |− 1 25 / 29
Trick C [ X, ( R ′ 2 )] =1 if and only if 1 , R ′ ∈ R ′ ∈ R ′ Y 1 2 X 25 / 29
Trick C [( R ′ 1 , R ′ 2 ) , Y ] =1 if and only if v Y ∈ R ′ Y ∈ R ′ 1 2 25 / 29
Trick C · C [ X, Y ] = 2 | CC ( X ∪ Y ) |− 1 GF (2) C · C = M 25 / 29
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