More applications of the d -neighbor equivalence: acyclic and - - PowerPoint PPT Presentation
More applications of the d -neighbor equivalence: acyclic and connectivity constraints Benjamin Bergougnoux , Mamadou Moustapha Kant IRIF, CNRS, Universit Paris Diderot, France LIMOS, CNRS, Universit Clermont Auvergne, France
Benjamin Bergougnoux⋆, Mamadou Moustapha Kanté†
⋆ IRIF, CNRS, Université Paris Diderot, France † LIMOS, CNRS, Université Clermont Auvergne, France
ESA, Munich, October 11, 2019
▸ 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.
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Recursively decompose your graph...
We recursively cut the vertex set in two
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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.
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Width of a decomposition D ∶= max f(cut) among the cuts of D.
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Width of a graph G ∶= min width of the decompositions of G.
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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) ≤ 2mw(G).
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Rank-width
Defined from the function rw(A) ∶= the rank of adjacency matrix between A and A over GF(2).
A A
1 1 1 1 1 1
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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
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tree-width clique-width rank-width mim-width
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▸ NP-hard for all these widths measures. ▸ Efficient algorithms for tree-width and rank-width. → Running time: 2O(k) ⋅ nO(1). ▸ Tough open questions for clique-width and mim-width.
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tree-width clique-width rank-width mim-width
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S1 S2 S
Combination Reduction
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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) ⋅ nO(1).
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Theorem [B., Kanté 2019]
We have a meta-algorithm for the connected and acyclic variants of (σ,ρ)-dominating set problems. ▸ Connected dominating set, ▸ Connected vertex cover, ▸ Node-weighted Steiner tree, ▸ Feedback vertex set, ▸ Maximum induced tree, ▸ Longest induced path, ▸ Maximum induced linear forest, ▸ Max. induced tree of ∆ ≤ 42,...
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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: tree-width 2O(tw) ⋅ nO(1) clique-width 2O(cw) ⋅ nO(1) rank-width 2O(rw2) ⋅ nO(1) mim-width nO(mim)
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necd(A): # of equivalence classes of the d-neighbor equivalence over A.
Theorem [Bui-Xuan, Telle, Vatshelle 2013]
It is enough to keep necd(A) ⋅ necd(A) partial solutions at each cut (A,A), for any (σ,ρ)-Dominating Set problem.
Lemma [Vatshelle 2012]
necd(A) is upper bounded by: tree-width 2d⋅tw ⋅ nO(1) clique-width 2d⋅cw ⋅ nO(1) rank-width 2d⋅rw2 ⋅ nO(1) mim-width nd⋅mim
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necd(A): # of equivalence classes of the d-neighbor equivalence over A.
Theorem [Bui-Xuan, Telle, Vatshelle 2013]
It is enough to keep necd(A) ⋅ necd(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: tree-width 2O(tw) ⋅ nO(1) clique-width 2O(cw) ⋅ nO(1) rank-width 2O(rw2) ⋅ nO(1) mim-width nO(mim)
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Theorem [B., Kanté 2019]
It is enough to keep necd(A) ⋅ necd(A) ⋅ nec1(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: tree-width 2O(tw) ⋅ nO(1) clique-width 2O(cw) ⋅ nO(1) rank-width 2O(rw2) ⋅ nO(1) mim-width nO(mim)
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Connected dominating set
Finding a vertex set D of minimum weight which dominates all the vertices and which induces a connected graph.
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1-neighbor equivalence relation
X,Y ⊆ A are 1-neighbor equivalent in A if N(X) ∩ A = N(Y ) ∩ A.
A X Y A
nec1(A) ∶= # of equivalence classes over A.
Theorem [B. and Kanté 2018]
It is enough to keep nec1(A)3nec1(A) partial solutions for each cut (A,A) to solve Connected dominating set.
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A set of partial solutions for each equivalence class R′ of the 1- neighbor equivalence over A.
Y
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The sets Y ∈ R′ have the same neighborhood in A.
Y
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Y ′
The sets Y ∈ R′ have the same neighborhood in A.
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X
Partial solutions associated with R′ : → X ⊆ A such that X ∪ Y dominates A, for (all) Y ∈ R′
Y
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Partial solutions associated with R′ : → X ⊆ A such that X ∪ Y dominates A, for (all) Y ∈ R′
Y X′
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At the root of the decomposition, the cut is (V (G), ∅) : → A partial solution is a dominating set.
X
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Y
Two partial solutions associated with R′ are equivalent for the dom- ination if they are 1-neighbor equivalent!
X
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Y
Two partial solutions associated with R′ are equivalent for the dom- ination if they are 1-neighbor equivalent!
X′
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Lemma [Bui-Xuan, Telle and Vatshelle, 2013]
For Dominating set, it is enough to keep nec1(A) ⋅ nec1(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
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▸ We need an equivalence relation between sets of partial solutions.
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▸ 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 ).
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▸ 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}.
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▸ 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 ⊆ 2A. ▸ Output: S⋆ ⊆ S such that ∣S⋆∣ ≤ nec1(A)2 and S⋆R′-represents S. ▸ Running time: ∣S∣ ⋅ nec1(A)O(1) ⋅ n2.
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Inspiration [Bodlaender, Cygan, Kratsch, Nederlof 2013]
Rank based approach: technique to obtained 2O(tw) ⋅ n time algorithms for many connectivity problems. Let M be the (S,R′)-matrix: M[X,Y ] ∶= ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ 1 if G[X ∪ Y ] is connected,
▸ 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.
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P: the set of pairs (R′
1,R′ 2) of 1-neighbor equivalence classes in A.
R′ S
R′ S
P
P
GF(2)
▸ A basis of C is also a basis of M. ▸ ∣P∣ = nec1(A)2. ▸ C is computable in time ∣S∣ ⋅ ∣P∣ ⋅ n2.
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vY There is 2|CC(X∪Y )|−1 ways of dividing X ∪ Y such that vY is on
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vY There is 2|CC(X∪Y )|−1 ways of dividing X ∪ Y such that vY is on
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1, R′ 2)]=1 if and only if
1
2
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1, R′ 2), Y ]=1 if and only if
1
2
vY
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GF(2)
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Theorem [B., Kanté 2019]
It is enough to keep necd(A) ⋅ necd(A) ⋅ nec1(A)2 ⋅ X partial solutions at each cut (A,A), for any acyclic and acyclic+connected variant of a (σ,ρ)-Dominating Set problem.
Corollary [B., Kanté 2019]
We can solve these problems in time: tree-width 2O(tw) ⋅ nO(1) clique-width 2O(cw) ⋅ nO(1) rank-width 2O(rw2) ⋅ nO(1) mim-width nO(mim)
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Thanks to the d-neighbor equivalence, we obtain the best algorithms: ▸ for many problems → (σ,ρ)-Dominating Set problems and their variants. ▸ for many width-measures → tree-width, clique-width, rank-width, mim-width, Q-rank-width. The algorithms for clique-width and tree-width are optimal under ETH. → What about rank-width: is 2O(rw2) ⋅ nO(1) optimal?
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Can we use the d-neighbor equivalence for W[1]-hard problems parameterized by clique-width? ▸ We can solve these 3 problems in time 2O(tw) ⋅ n and nO(cw). → Optimal under ETH. ▸ Using the d-neighbor equivalence width d a constant is useless.
Theorem [B., Kanté 19]
For Max Cut, it is enough to keep necn(A) partial solutions at each cut (A,A).
Corollary [B., Kanté 19]
We can solve Max Cut in time nO(cw), nO(rwQ) and n2O(rw).
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Towards a “Courcelle’s theorem” which gives efficient algorithms?
A problem Algorithm for tw Algorithm for cw Constraints basket Best Best
Example:
We want a set of vertices X of minimum weight satisfying:
DominatingSet(X) ∧ Acyclic(X) ∧ Connected(X).
Thm ⇒ We can find one in time: 2O(tw) ⋅ n, 2O(cw) ⋅ n, 2O(rw2) ⋅ n3.
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