Subexponential parameterized complexity of completion problems Survey of the upper bounds
Marcin Pilipczuk
University of Warsaw
4th Nov 2015
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 1/22 Theory: result should have some property. Modification: error in our measurments. A few errors ⇒ #modifications as a parameter.
Graph modification problems
An input graph: a result of an experiment.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 2/22 Modification: error in our measurments. A few errors ⇒ #modifications as a parameter.
Graph modification problems
An input graph: a result of an experiment. Theory: result should have some property.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 2/22 A few errors ⇒ #modifications as a parameter.
Graph modification problems
An input graph: a result of an experiment. Theory: result should have some property. Modification: error in our measurments.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 2/22 Graph modification problems
An input graph: a result of an experiment. Theory: result should have some property. Modification: error in our measurments. A few errors ⇒ #modifications as a parameter.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 2/22 G Completion F-Completion obtain a graph ∈ G kill all induced subgraphs ∈ F
Side note: no known P vs NP dichotomy for completion problems.
Completions
completion = only edge insertions are allowed
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 3/22 F-Completion kill all induced subgraphs ∈ F
Side note: no known P vs NP dichotomy for completion problems.
Completions
completion = only edge insertions are allowed
G Completion obtain a graph ∈ G
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 3/22 Side note: no known P vs NP dichotomy for completion problems.
Completions
completion = only edge insertions are allowed
G Completion F-Completion obtain a graph ∈ G kill all induced subgraphs ∈ F
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 3/22 Completions
completion = only edge insertions are allowed
G Completion F-Completion obtain a graph ∈ G kill all induced subgraphs ∈ F
Side note: no known P vs NP dichotomy for completion problems.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 3/22 Interesting classes for completion
threshold vertex cover ⊆ ≥
proper trivially perfect treedepth bandwidth interval
≥ ≤ ⊆ ⊇
pathwidth interval ≥ ⊆
chordal treewidth
measure(G) ' min{ω(H): H ∈ G and H is a completion of G}.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 4/22 Disclaimer
In this talk we mostly focus on f (k) in the FPT running time f (k)nO(1).
We denote O∗(f (k)) = f (k)nO(1).
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 5/22 Theorem (Cai, IPL’96) A simple branching strategy ⇒ O∗(ck ) FPT algorithm
Works also for: Co-cluster, Cograph, Threshold, Pseudosplit,...
Finite set F
Split Completion {2K2, C4, C5}-Completion
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 6/22 Works also for: Co-cluster, Cograph, Threshold, Pseudosplit,...
Finite set F
Split Completion {2K2, C4, C5}-Completion
Theorem (Cai, IPL’96) A simple branching strategy ⇒ O∗(ck ) FPT algorithm
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 6/22 Finite set F
Split Completion {2K2, C4, C5}-Completion
Theorem (Cai, IPL’96) A simple branching strategy ⇒ O∗(ck ) FPT algorithm
Works also for: Co-cluster, Cograph, Threshold, Pseudosplit,...
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 6/22 Theorem (Kaplan, Shamir, Tarjan, SICOMP’99) Large hole ⇒ many options but big cost (Catalan number C`−2 for ` − 3 edges) ⇒ O∗(4k ) branching.
Gives also O∗(ck ) FPT algorithm for Proper Interval Completion, as Chordal −→ Proper Interval means killing {S3, claw, net}.
Chordal completion
Chordal Completion {C4, C5, C6,...}-Completion
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 7/22 Theorem (Kaplan, Shamir, Tarjan, SICOMP’99) Large hole ⇒ many options but big cost (Catalan number C`−2 for ` − 3 edges) ⇒ O∗(4k ) branching.
Gives also O∗(ck ) FPT algorithm for Proper Interval Completion, as Chordal −→ Proper Interval means killing {S3, claw, net}.
Chordal completion
Chordal Completion {C4, C5, C6,...}-Completion
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 7/22 Theorem (Kaplan, Shamir, Tarjan, SICOMP’99) Large hole ⇒ many options but big cost (Catalan number C`−2 for ` − 3 edges) ⇒ O∗(4k ) branching.
Gives also O∗(ck ) FPT algorithm for Proper Interval Completion, as Chordal −→ Proper Interval means killing {S3, claw, net}.
Chordal completion
Chordal Completion {C4, C5, C6,...}-Completion
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 7/22 Theorem (Kaplan, Shamir, Tarjan, SICOMP’99) Large hole ⇒ many options but big cost (Catalan number C`−2 for ` − 3 edges) ⇒ O∗(4k ) branching.
Gives also O∗(ck ) FPT algorithm for Proper Interval Completion, as Chordal −→ Proper Interval means killing {S3, claw, net}.
Chordal completion
Chordal Completion {C4, C5, C6,...}-Completion
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 7/22 Gives also O∗(ck ) FPT algorithm for Proper Interval Completion, as Chordal −→ Proper Interval means killing {S3, claw, net}.
Chordal completion
Chordal Completion {C4, C5, C6,...}-Completion
Theorem (Kaplan, Shamir, Tarjan, SICOMP’99) Large hole ⇒ many options but big cost (Catalan number C`−2 for ` − 3 edges) ⇒ O∗(4k ) branching.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 7/22 Chordal completion
Chordal Completion {C4, C5, C6,...}-Completion
Theorem (Kaplan, Shamir, Tarjan, SICOMP’99) Large hole ⇒ many options but big cost (Catalan number C`−2 for ` − 3 edges) ⇒ O∗(4k ) branching.
Gives also O∗(ck ) FPT algorithm for Proper Interval Completion, as Chordal −→ Proper Interval means killing {S3, claw, net}.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 7/22 Long-standing open problem for more than a decade.
Theorem (Villanger, Heggernes, Paul, Telle, SICOMP’09) Branching still doable! An O∗(k2k ) FPT algorithm.
Theorem (Cao, SODA’16) Can be solved in O(ck (n + m)) time.
Interval completion
Interval Completion {holes, ATs}-Completion
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 8/22 Long-standing open problem for more than a decade.
Theorem (Villanger, Heggernes, Paul, Telle, SICOMP’09) Branching still doable! An O∗(k2k ) FPT algorithm.
Theorem (Cao, SODA’16) Can be solved in O(ck (n + m)) time.
Interval completion
Interval Completion {holes, ATs}-Completion
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 8/22 Theorem (Villanger, Heggernes, Paul, Telle, SICOMP’09) Branching still doable! An O∗(k2k ) FPT algorithm.
Theorem (Cao, SODA’16) Can be solved in O(ck (n + m)) time.
Interval completion
Interval Completion {holes, ATs}-Completion
Long-standing open problem for more than a decade.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 8/22 Interval completion
Interval Completion {holes, ATs}-Completion
Long-standing open problem for more than a decade.
Theorem (Villanger, Heggernes, Paul, Telle, SICOMP’09) Branching still doable! An O∗(k2k ) FPT algorithm.
Theorem (Cao, SODA’16) Can be solved in O(ck (n + m)) time.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 8/22 Theorem (Guo, ISAAC’07) Chain, Split, Threshold and Trivially Perfect Completion admit polynomial kernels.
Theorem (Kratsch, Wahlstr¨om,IWPEC’09) There is a finite F such that F-Completion does not admit a polynomial kernel unless NP ⊆ coNP/poly.
Theorem (Guillemot, Havet, Paul, Perez, Algorithmica’13) Cograph Completion admits a polynomial kernel.
Theorem (Bessy, Perez, Inf. Comp.’13) Proper Interval Completion admits a polynomial kernel.
Polynomial kernels
Theorem (Kaplan, Shamir, Tarjan, SICOMP’99) Chordal Completion admits a polynomial kernel.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 9/22 Theorem (Kratsch, Wahlstr¨om,IWPEC’09) There is a finite F such that F-Completion does not admit a polynomial kernel unless NP ⊆ coNP/poly.
Theorem (Guillemot, Havet, Paul, Perez, Algorithmica’13) Cograph Completion admits a polynomial kernel.
Theorem (Bessy, Perez, Inf. Comp.’13) Proper Interval Completion admits a polynomial kernel.
Polynomial kernels
Theorem (Kaplan, Shamir, Tarjan, SICOMP’99) Chordal Completion admits a polynomial kernel.
Theorem (Guo, ISAAC’07) Chain, Split, Threshold and Trivially Perfect Completion admit polynomial kernels.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 9/22 Theorem (Guillemot, Havet, Paul, Perez, Algorithmica’13) Cograph Completion admits a polynomial kernel.
Theorem (Bessy, Perez, Inf. Comp.’13) Proper Interval Completion admits a polynomial kernel.
Polynomial kernels
Theorem (Kaplan, Shamir, Tarjan, SICOMP’99) Chordal Completion admits a polynomial kernel.
Theorem (Guo, ISAAC’07) Chain, Split, Threshold and Trivially Perfect Completion admit polynomial kernels.
Theorem (Kratsch, Wahlstr¨om,IWPEC’09) There is a finite F such that F-Completion does not admit a polynomial kernel unless NP ⊆ coNP/poly.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 9/22 Polynomial kernels
Theorem (Kaplan, Shamir, Tarjan, SICOMP’99) Chordal Completion admits a polynomial kernel.
Theorem (Guo, ISAAC’07) Chain, Split, Threshold and Trivially Perfect Completion admit polynomial kernels.
Theorem (Kratsch, Wahlstr¨om,IWPEC’09) There is a finite F such that F-Completion does not admit a polynomial kernel unless NP ⊆ coNP/poly.
Theorem (Guillemot, Havet, Paul, Perez, Algorithmica’13) Cograph Completion admits a polynomial kernel.
Theorem (Bessy, Perez, Inf. Comp.’13) Proper Interval Completion admits a polynomial kernel.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 9/22 Fomin, Villanger, SODA’12: think positively!
Theorem (Fomin, Villanger, SODA’12) √ Chordal Completion can be solved in time O∗(kO( k)).
Look at the bright side!
G Completion F-Completion
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 10/22 Theorem (Fomin, Villanger, SODA’12) √ Chordal Completion can be solved in time O∗(kO( k)).
Look at the bright side!
G Completion F-Completion
Fomin, Villanger, SODA’12: think positively!
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 10/22 Look at the bright side!
G Completion F-Completion
Fomin, Villanger, SODA’12: think positively!
Theorem (Fomin, Villanger, SODA’12) √ Chordal Completion can be solved in time O∗(kO( k)).
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 10/22 crucial structure√ ' a maximal clique; often nO( k) candidate structures; with poly kernel ⇒ 2o(k) bound
1 Apply known polynomial kernel. 2 Identify and enumerate 2o(k) candidates for crucial structures in G.
3 Candidate structure −→ DP state.
Build the structure of (G + completion) by dynamic programming. Key observation: there are 2o(k) reasonable ‘partial structures’, and hence 2o(k) reasonable DP states. Strategy:
Look at the bright side!
Fomin, Villanger, SODA’12: think positively!
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 11/22 crucial structure√ ' a maximal clique; often nO( k) candidate structures; with poly kernel ⇒ 2o(k) bound
1 Apply known polynomial kernel. 2 Identify and enumerate 2o(k) candidates for crucial structures in G.
3 Candidate structure −→ DP state.
Key observation: there are 2o(k) reasonable ‘partial structures’, and hence 2o(k) reasonable DP states. Strategy:
Look at the bright side!
Fomin, Villanger, SODA’12: think positively!
Build the structure of (G + completion) by dynamic programming.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 11/22 crucial structure√ ' a maximal clique; often nO( k) candidate structures; with poly kernel ⇒ 2o(k) bound
1 Apply known polynomial kernel. 2 Identify and enumerate 2o(k) candidates for crucial structures in G.
3 Candidate structure −→ DP state.
Strategy:
Look at the bright side!
Fomin, Villanger, SODA’12: think positively!
Build the structure of (G + completion) by dynamic programming. Key observation: there are 2o(k) reasonable ‘partial structures’, and hence 2o(k) reasonable DP states.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 11/22 crucial structure√ ' a maximal clique; often nO( k) candidate structures; with poly kernel ⇒ 2o(k) bound
1 Apply known polynomial kernel. 2 Identify and enumerate 2o(k) candidates for crucial structures in G.
3 Candidate structure −→ DP state.
Look at the bright side!
Fomin, Villanger, SODA’12: think positively!
Build the structure of (G + completion) by dynamic programming. Key observation: there are 2o(k) reasonable ‘partial structures’, and hence 2o(k) reasonable DP states. Strategy:
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 11/22 crucial structure√ ' a maximal clique; often nO( k) candidate structures; with poly kernel ⇒ 2o(k) bound
2 Identify and enumerate 2o(k) candidates for crucial structures in G.
3 Candidate structure −→ DP state.
Look at the bright side!
Fomin, Villanger, SODA’12: think positively!
Build the structure of (G + completion) by dynamic programming. Key observation: there are 2o(k) reasonable ‘partial structures’, and hence 2o(k) reasonable DP states. Strategy:
1 Apply known polynomial kernel.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 11/22 crucial structure√ ' a maximal clique; often nO( k) candidate structures; with poly kernel ⇒ 2o(k) bound
3 Candidate structure −→ DP state.
Look at the bright side!
Fomin, Villanger, SODA’12: think positively!
Build the structure of (G + completion) by dynamic programming. Key observation: there are 2o(k) reasonable ‘partial structures’, and hence 2o(k) reasonable DP states. Strategy:
1 Apply known polynomial kernel. 2 Identify and enumerate 2o(k) candidates for crucial structures in G.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 11/22 √ often nO( k) candidate structures; with poly kernel ⇒ 2o(k) bound
3 Candidate structure −→ DP state.
Look at the bright side!
Fomin, Villanger, SODA’12: think positively!
Build the structure of (G + completion) by dynamic programming. Key observation: there are 2o(k) reasonable ‘partial structures’, and hence 2o(k) reasonable DP states. Strategy:
1 Apply known polynomial kernel. 2 Identify and enumerate 2o(k) candidates for crucial structures in G. crucial structure ' a maximal clique;
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 11/22 3 Candidate structure −→ DP state.
Look at the bright side!
Fomin, Villanger, SODA’12: think positively!
Build the structure of (G + completion) by dynamic programming. Key observation: there are 2o(k) reasonable ‘partial structures’, and hence 2o(k) reasonable DP states. Strategy:
1 Apply known polynomial kernel. 2 Identify and enumerate 2o(k) candidates for crucial structures in G.
crucial structure√ ' a maximal clique; often nO( k) candidate structures; with poly kernel ⇒ 2o(k) bound
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 11/22 Look at the bright side!
Fomin, Villanger, SODA’12: think positively!
Build the structure of (G + completion) by dynamic programming. Key observation: there are 2o(k) reasonable ‘partial structures’, and hence 2o(k) reasonable DP states. Strategy:
1 Apply known polynomial kernel. 2 Identify and enumerate 2o(k) candidates for crucial structures in G.
crucial structure√ ' a maximal clique; often nO( k) candidate structures; with poly kernel ⇒ 2o(k) bound
3 Candidate structure −→ DP state.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 11/22 √ O∗ kO( k)
√ 2/3 O∗ kO( k) O∗ kO(k )
√ O∗ kO( k)
√ O∗ kO( k)
Look at the bright side!
threshold ⊆
trivially perfect proper interval
⊆ ⊇
interval ⊆
chordal
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 12/22 Look at the bright side!
threshold √ O∗ kO( k) ⊆
trivially perfect proper interval √ 2/3 O∗ kO( k) O∗ kO(k )
⊆ ⊇
interval √ O∗ kO( k) ⊆
chordal √ O∗ kO( k)
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 12/22 / minimal clique separator
crucial structure = maximal clique
Theorem (Fomin, Villanger, SODA’12) √ O( k) One can either enumerate k candidate maximal√ cliques and candidate clique separators, or perform a kO( k) branching.
DP state = clique separator Ω+ one connected component C of G \ Ω, value = minimum completion of G[C ∪ Ω] that cliquifies Ω.
√ Corollary: Chain Completion in O∗(kO( k)) time.
Chordal completion
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 13/22 / minimal clique separator
Theorem (Fomin, Villanger, SODA’12) √ O( k) One can either enumerate k candidate maximal√ cliques and candidate clique separators, or perform a kO( k) branching.
DP state = clique separator Ω+ one connected component C of G \ Ω, value = minimum completion of G[C ∪ Ω] that cliquifies Ω.
√ Corollary: Chain Completion in O∗(kO( k)) time.
Chordal completion
crucial structure = maximal clique
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 13/22 Theorem (Fomin, Villanger, SODA’12) √ O( k) One can either enumerate k candidate maximal√ cliques and candidate clique separators, or perform a kO( k) branching.
DP state = clique separator Ω+ one connected component C of G \ Ω, value = minimum completion of G[C ∪ Ω] that cliquifies Ω.
√ Corollary: Chain Completion in O∗(kO( k)) time.
Chordal completion
crucial structure = maximal clique / minimal clique separator
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 13/22 DP state = clique separator Ω+ one connected component C of G \ Ω, value = minimum completion of G[C ∪ Ω] that cliquifies Ω.
√ Corollary: Chain Completion in O∗(kO( k)) time.
Chordal completion
crucial structure = maximal clique / minimal clique separator
Theorem (Fomin, Villanger, SODA’12) √ O( k) One can either enumerate k candidate maximal√ cliques and candidate clique separators, or perform a kO( k) branching.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 13/22 √ Corollary: Chain Completion in O∗(kO( k)) time.
Chordal completion
crucial structure = maximal clique / minimal clique separator
Theorem (Fomin, Villanger, SODA’12) √ O( k) One can either enumerate k candidate maximal√ cliques and candidate clique separators, or perform a kO( k) branching.
DP state = clique separator Ω+ one connected component C of G \ Ω, value = minimum completion of G[C ∪ Ω] that cliquifies Ω.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 13/22 Chordal completion
crucial structure = maximal clique / minimal clique separator
Theorem (Fomin, Villanger, SODA’12) √ O( k) One can either enumerate k candidate maximal√ cliques and candidate clique separators, or perform a kO( k) branching.
DP state = clique separator Ω+ one connected component C of G \ Ω, value = minimum completion of G[C ∪ Ω] that cliquifies Ω.
√ Corollary: Chain Completion in O∗(kO( k)) time.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 13/22 Multiple-step, involved combinatorial analysis. Will give a flavour on the Interval Completion case.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are nO( k) reasonable candidates for maximal cliques in the Interval Completion case.
Enumerating potential maximal cliques
Theorem (Fomin, Villanger, SODA’12) √ O( k) One can either enumerate k candidate maximal√ cliques and candidate clique separators, or perform a kO( k) branching.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 14/22 Will give a flavour on the Interval Completion case.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are nO( k) reasonable candidates for maximal cliques in the Interval Completion case.
Enumerating potential maximal cliques
Theorem (Fomin, Villanger, SODA’12) √ O( k) One can either enumerate k candidate maximal√ cliques and candidate clique separators, or perform a kO( k) branching.
Multiple-step, involved combinatorial analysis.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 14/22 Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are nO( k) reasonable candidates for maximal cliques in the Interval Completion case.
Enumerating potential maximal cliques
Theorem (Fomin, Villanger, SODA’12) √ O( k) One can either enumerate k candidate maximal√ cliques and candidate clique separators, or perform a kO( k) branching.
Multiple-step, involved combinatorial analysis. Will give a flavour on the Interval Completion case.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 14/22 Enumerating potential maximal cliques
Theorem (Fomin, Villanger, SODA’12) √ O( k) One can either enumerate k candidate maximal√ cliques and candidate clique separators, or perform a kO( k) branching.
Multiple-step, involved combinatorial analysis. Will give a flavour on the Interval Completion case.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are nO( k) reasonable candidates for maximal cliques in the Interval Completion case.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 14/22 Ω v2 v1 c1 c2 x y
√ √ at most 2 k right ends at most 2 k left ends call them $1 call them $2 √ cheap v = at most k incident solution edges. c1 := last ending cheap before Ω c2 := first starting cheap after Ω Why x ∈ Ω?
x has left end before Ω because some y ∈ NG (x) has right end before Ω.
y ∈ NG (x) for some y with right end before Ω ⇒ x ∈ NG (v1) ∪ NG ($1) ∪ NG+F (c1).
Maximal cliques in Interval Completion
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 15/22 c1 c2 x y
√ √ at most 2 k right ends at most 2 k left ends call them $1 call them $2 √ cheap v = at most k incident solution edges. c1 := last ending cheap before Ω c2 := first starting cheap after Ω Why x ∈ Ω?
x has left end before Ω because some y ∈ NG (x) has right end before Ω.
y ∈ NG (x) for some y with right end before Ω ⇒ x ∈ NG (v1) ∪ NG ($1) ∪ NG+F (c1).
Maximal cliques in Interval Completion
Ω v2 v1
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 15/22 x y
√ √ at most 2 k right ends at most 2 k left ends call them $1 call them $2
Why x ∈ Ω?
x has left end before Ω because some y ∈ NG (x) has right end before Ω.
y ∈ NG (x) for some y with right end before Ω ⇒ x ∈ NG (v1) ∪ NG ($1) ∪ NG+F (c1).
Maximal cliques in Interval Completion
Ω v2 v1 c1 c2
√ cheap v = at most k incident solution edges. c1 := last ending cheap before Ω c2 := first starting cheap after Ω
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 15/22 x y
Why x ∈ Ω?
x has left end before Ω because some y ∈ NG (x) has right end before Ω.
y ∈ NG (x) for some y with right end before Ω ⇒ x ∈ NG (v1) ∪ NG ($1) ∪ NG+F (c1).
Maximal cliques in Interval Completion
Ω v2 v1 c1 c2
√ √ at most 2 k right ends at most 2 k left ends call them $1 call them $2 √ cheap v = at most k incident solution edges. c1 := last ending cheap before Ω c2 := first starting cheap after Ω
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 15/22 y ∈ NG (x) for some y with right end before Ω ⇒ x ∈ NG (v1) ∪ NG ($1) ∪ NG+F (c1).
Maximal cliques in Interval Completion
Ω v2 v1 c1 c2 x y
√ √ at most 2 k right ends at most 2 k left ends call them $1 call them $2 √ cheap v = at most k incident solution edges. c1 := last ending cheap before Ω c2 := first starting cheap after Ω Why x ∈ Ω?
x has left end before Ω because some y ∈ NG (x) has right end before Ω.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 15/22 Maximal cliques in Interval Completion
Ω v2 v1 c1 c2 x y
√ √ at most 2 k right ends at most 2 k left ends call them $1 call them $2 √ cheap v = at most k incident solution edges. c1 := last ending cheap before Ω c2 := first starting cheap after Ω Why x ∈ Ω?
x has left end before Ω because some y ∈ NG (x) has right end before Ω.
y ∈ NG (x) for some y with right end before Ω ⇒ x ∈ NG (v1) ∪ NG ($1) ∪ NG+F (c1).
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 15/22 √ There are nO( k) choices for: v , v , c , c ; 1 2 1 2 √ $1 and $1, as there are of size O( k);
solution edges incident to c1 and c2, as both c1 and c2 are cheap.
Maximal cliques in Interval Completion
Ω v2 v1 c1 c2 x y
√ √ at most 2 k right ends at most 2 k left ends call them $1 call them $2
Lemma
Ω = (NG [v1] ∪ NG ($1) ∪ NG+F (c1)) ∩ (NG [v2] ∪ NG ($2) ∪ NG+F (c2)).
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 16/22 Maximal cliques in Interval Completion
Ω v2 v1 c1 c2 x y
√ √ at most 2 k right ends at most 2 k left ends call them $1 call them $2
Lemma
Ω = (NG [v1] ∪ NG ($1) ∪ NG+F (c1)) ∩ (NG [v2] ∪ NG ($2) ∪ NG+F (c2)). √ There are nO( k) choices for: v , v , c , c ; 1 2 1 2 √ $1 and $1, as there are of size O( k);
solution edges incident to c1 and c2, as both c1 and c2 are cheap.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 16/22 ? or we can reduce something.
Problem 1: no known polynomial kernel!
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are nO( k) reasonable candidates for maximal cliques.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ For any vertex v, there are kO(t+ k)nO(1) reasonable ways to choose completion edges incident to v, as long as there are at most t of them?.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are kO( k)n8 reasonable candidates for maximal cliques?.
Interval Completion
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 17/22 ? or we can reduce something.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are nO( k) reasonable candidates for maximal cliques.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ For any vertex v, there are kO(t+ k)nO(1) reasonable ways to choose completion edges incident to v, as long as there are at most t of them?.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are kO( k)n8 reasonable candidates for maximal cliques?.
Interval Completion
Problem 1: no known polynomial kernel!
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 17/22 ? or we can reduce something.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are nO( k) reasonable candidates for maximal cliques.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ For any vertex v, there are kO(t+ k)nO(1) reasonable ways to choose completion edges incident to v, as long as there are at most t of them?.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are kO( k)n8 reasonable candidates for maximal cliques?.
Interval Completion
Problem 1: no known polynomial kernel!
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 17/22 ? or we can reduce something.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ For any vertex v, there are kO(t+ k)nO(1) reasonable ways to choose completion edges incident to v, as long as there are at most t of them?.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are kO( k)n8 reasonable candidates for maximal cliques?.
Interval Completion
Problem 1: no known polynomial kernel!
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are nO( k) reasonable candidates for maximal cliques.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 17/22 Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are kO( k)n8 reasonable candidates for maximal cliques?.
? or we can reduce something.
Interval Completion
Problem 1: no known polynomial kernel!
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are nO( k) reasonable candidates for maximal cliques.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ For any vertex v, there are kO(t+ k)nO(1) reasonable ways to choose completion edges incident to v, as long as there are at most t of them?.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 17/22 Interval Completion
Problem 1: no known polynomial kernel!
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are nO( k) reasonable candidates for maximal cliques.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ For any vertex v, there are kO(t+ k)nO(1) reasonable ways to choose completion edges incident to v, as long as there are at most t of them?.
Theorem (Bliznets, Fomin, P., Pilipczuk, 2014) √ There are kO( k)n8 reasonable candidates for maximal cliques?.
? or we can reduce something.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 17/22 Solution: make much more complicated DP states.
Interval Completion
Problem 2: history is hard to deduce!
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 18/22 Solution: make much more complicated DP states.
Interval Completion
Problem 2: history is hard to deduce!
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 18/22 Solution: make much more complicated DP states.
Interval Completion
Problem 2: history is hard to deduce!
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 18/22 Solution: make much more complicated DP states.
Interval Completion
Problem 2: history is hard to deduce!
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 18/22 Solution: make much more complicated DP states.
Interval Completion
Problem 2: history is hard to deduce!
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 18/22 Interval Completion
Problem 2: history is hard to deduce! Solution: make much more complicated DP states.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 18/22 Theorem (Drange, Fomin, Pilipczuk, Villanger, STACS’14) √ O∗(kO( k)) algorithms for: 1 Trivially Perfect Completion, 2 Threshold Completion via chromatic coding + reduction to Chain Completion, 3 Pseudosplit Completion similarly as Split Completion.
More completions
Theorem (Ghosh, Kolay, Kumar, Misra, Panolan, Rai, Ramanujan, SWAT’12) √ An O∗(kO( k)) algorithm for Split Completion via chromatic coding.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 19/22 More completions
Theorem (Ghosh, Kolay, Kumar, Misra, Panolan, Rai, Ramanujan, SWAT’12) √ An O∗(kO( k)) algorithm for Split Completion via chromatic coding.
Theorem (Drange, Fomin, Pilipczuk, Villanger, STACS’14) √ O∗(kO( k)) algorithms for: 1 Trivially Perfect Completion, 2 Threshold Completion via chromatic coding + reduction to Chain Completion, 3 Pseudosplit Completion similarly as Split Completion.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 19/22 Co-cluster ETH-hard [KU, DAM’12] Cograph ETH-hard [DMPV, STACS’14] Co-Trivially Perfect ETH-hard [DMPV, STACS’14]
Summary
√ O∗(kO( k)) [FV, SODA’12] Chordal √ O∗(kO( k)) [FV, SODA’12] Chain √ O∗(kO( k)) [GKKMPRR, SWAT’12] Split √ O∗(kO( k)) [DMPV, STACS’14] Trivially Perfect √ O∗(kO( k)) [DMPV, STACS’14] Threshold √ O∗(kO( k)) [DMPV, STACS’14] Pseudosplit √ Interval O∗(kO( k)) [BFPP, SODA’16] 2/3 Proper Interval O∗(kO(k )) [BFPP, ESA’14]
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 20/22 Summary
√ O∗(kO( k)) [FV, SODA’12] Chordal √ O∗(kO( k)) [FV, SODA’12] Chain √ O∗(kO( k)) [GKKMPRR, SWAT’12] Split √ O∗(kO( k)) [DMPV, STACS’14] Trivially Perfect √ O∗(kO( k)) [DMPV, STACS’14] Threshold √ O∗(kO( k)) [DMPV, STACS’14] Pseudosplit √ Interval O∗(kO( k)) [BFPP, SODA’16] 2/3 Proper Interval O∗(kO(k )) [BFPP, ESA’14] Co-cluster ETH-hard [KU, DAM’12] Cograph ETH-hard [DMPV, STACS’14] Co-Trivially Perfect ETH-hard [DMPV, STACS’14]
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 20/22 Does Interval Completion admit a polynomial kernel? √ Can you provide 2Ω( k) lower bounds under ETH? √ An O∗(2O( k))-time algorithm for one of the problems? √ An O∗(2o( k))-time algorithm seems hard, as it gives 2o(n) bound. Is there any meta-explanation why there are subexponential algorithms for this family of problems?
Open problems
√ Can you get time O∗(kO( k)) for Proper Interval Completion?
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 21/22 √ Can you provide 2Ω( k) lower bounds under ETH? √ An O∗(2O( k))-time algorithm for one of the problems? √ An O∗(2o( k))-time algorithm seems hard, as it gives 2o(n) bound. Is there any meta-explanation why there are subexponential algorithms for this family of problems?
Open problems
√ Can you get time O∗(kO( k)) for Proper Interval Completion? Does Interval Completion admit a polynomial kernel?
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 21/22 Is there any meta-explanation why there are subexponential algorithms for this family of problems?
Open problems
√ Can you get time O∗(kO( k)) for Proper Interval Completion? Does Interval Completion admit a polynomial kernel? √ Can you provide 2Ω( k) lower bounds under ETH? √ An O∗(2O( k))-time algorithm for one of the problems? √ An O∗(2o( k))-time algorithm seems hard, as it gives 2o(n) bound.
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 21/22 Open problems
√ Can you get time O∗(kO( k)) for Proper Interval Completion? Does Interval Completion admit a polynomial kernel? √ Can you provide 2Ω( k) lower bounds under ETH? √ An O∗(2O( k))-time algorithm for one of the problems? √ An O∗(2o( k))-time algorithm seems hard, as it gives 2o(n) bound. Is there any meta-explanation why there are subexponential algorithms for this family of problems?
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 21/22 Summary diagram
threshold √ O∗ kO( k) ⊆
trivially perfect proper interval √ 2/3 O∗ kO( k) O∗ kO(k )
⊆ ⊇
interval √ O∗ kO( k) Questions? ⊆
chordal √ O∗ kO( k)
Marcin Pilipczuk Subexponential parameterized complexity of completion problems 22/22