Polynomial kernels for constant-factor approximable problems

Transcription

1 1 Polynomial kernels for constant-factor approximable problems Stefan Kratsch November 11, 2010

2 2 What do these problems have in common? Cluster Edge Deletion, Cluster Edge Editing, Edge Dominating Set, Feedback Vertex Set, d-hitting Set, Maximum Cut, Maximum Knapsack, Maximum Leaf Spanning Tree, Maximum Satisfiability, Multicut in Trees, H-, Planar Dominating Set, Planar Independent Set, Set Splitting, Triangle Edge Deletion, Vertex Cover, H-free Vertex Deletion

3 2 What do these problems have in common? Cluster Edge Deletion, Cluster Edge Editing, Edge Dominating Set, Feedback Vertex Set, d-hitting Set, Maximum Cut, Maximum Knapsack, Maximum Leaf Spanning Tree, Maximum Satisfiability, Multicut in Trees, H-, Planar Dominating Set, Planar Independent Set, Set Splitting, Triangle Edge Deletion, Vertex Cover, H-free Vertex Deletion All these problems are constant-factor approximable and admit a polynomial kernel.

4 3 Relation between these classes? APX XP PTAS? FPT EPTAS PK FPTAS LK

5 4 EPTAS FPT If an optimization problem Q admits an EPTAS then its standard parameterization is FPT. [Bazgan 95]

6 4 EPTAS FPT If an optimization problem Q admits an EPTAS then its standard parameterization is FPT. [Bazgan 95] Proof sketch. Let A be an EPTAS for min-problem Q of runtime f ( 1 ɛ ) no(1). Given (x, k), decide opt(x) k?:

7 4 EPTAS FPT If an optimization problem Q admits an EPTAS then its standard parameterization is FPT. [Bazgan 95] Proof sketch. Let A be an EPTAS for min-problem Q of runtime f ( 1 ɛ ) no(1). Given (x, k), decide opt(x) k?: run A(x, ɛ) with ɛ = 1 k+1

8 4 EPTAS FPT If an optimization problem Q admits an EPTAS then its standard parameterization is FPT. [Bazgan 95] Proof sketch. Let A be an EPTAS for min-problem Q of runtime f ( 1 ɛ ) no(1). Given (x, k), decide opt(x) k?: run A(x, ɛ) with ɛ = 1 k+1 if opt(x) > k then A(x, ɛ) > k

9 4 EPTAS FPT If an optimization problem Q admits an EPTAS then its standard parameterization is FPT. [Bazgan 95] Proof sketch. Let A be an EPTAS for min-problem Q of runtime f ( 1 ɛ ) no(1). Given (x, k), decide opt(x) k?: run A(x, ɛ) with ɛ = 1 k+1 if opt(x) > k then A(x, ɛ) > k if opt(x) k then A(x, ɛ) (1 + 1 k+1 )k < k + 1

10 4 EPTAS FPT If an optimization problem Q admits an EPTAS then its standard parameterization is FPT. [Bazgan 95] Proof sketch. Let A be an EPTAS for min-problem Q of runtime f ( 1 ɛ ) no(1). Given (x, k), decide opt(x) k?: run A(x, ɛ) with ɛ = 1 k+1 if opt(x) > k then A(x, ɛ) > k if opt(x) k then A(x, ɛ) (1 + 1 k+1 )k < k + 1 thus A(x, ɛ) k

11 4 EPTAS FPT If an optimization problem Q admits an EPTAS then its standard parameterization is FPT. [Bazgan 95] Proof sketch. Let A be an EPTAS for min-problem Q of runtime f ( 1 ɛ ) no(1). Given (x, k), decide opt(x) k?: run A(x, ɛ) with ɛ = 1 k+1 if opt(x) > k then A(x, ɛ) > k if opt(x) k then A(x, ɛ) (1 + 1 k+1 )k < k + 1 thus A(x, ɛ) k It is crucial that the target function is integer-valued. (e.g., Vazirani s book defines it as rational)

12 5 What we know APX XP PTAS EPTAS FPT PK FPTAS LK We can run any approximation scheme with ɛ = 1 k+1 to get a parameterized algorithm with the same runtime.

13 6 FPTAS Linear kernel assume Q has an FPTAS with runtime α ( 1 ɛ ) β n γ then p-q can be solved in time α(k + 1) β n γ

14 6 FPTAS Linear kernel assume Q has an FPTAS with runtime α ( 1 ɛ ) β n γ then p-q can be solved in time α(k + 1) β n γ on input (x, k) run that algorithm for αn β+γ steps if it finishes then report the answer

15 6 FPTAS Linear kernel assume Q has an FPTAS with runtime α ( 1 ɛ ) β n γ then p-q can be solved in time α(k + 1) β n γ on input (x, k) run that algorithm for αn β+γ steps if it finishes then report the answer else α(k + 1) β n γ > αn β+γ

16 6 FPTAS Linear kernel assume Q has an FPTAS with runtime α ( 1 ɛ ) β n γ then p-q can be solved in time α(k + 1) β n γ on input (x, k) run that algorithm for αn β+γ steps if it finishes then report the answer else α(k + 1) β n γ > αn β+γ (k + 1) β > n β, and n O(k) also: eff-fpt LK

17 7 What we know APX XP PTAS EPTAS FPT PK FPTAS LK

18 7 What we know APX XP PTAS FPT EPTAS? PK FPTAS LK Bin APX, but NP-hard for k = 2 IS of unit disks PTAS, but W[1]-hard [Hunt et al. 98, Marx 05]

19 8 Where does that leave us? Complexity theory is intended to study the most general forms of computations. However the generality often poses a barrier, when one tries to formalize a noticeable trend among natural problems. Typically, the moment one tries to formalize the trend, a counterexample is found. [...] In contrast, specializing classes such as NP, PSPACE and NPO to Boolean constraint satisfaction creates miniaturized imprints of these classes that are refined enough to exhibit many phenomena over all problems contained in them. from Complexity Classifications of Boolean by Creignou et al.

20 9 Outline

21 10 Outline

22 11 Max SNP A problem Q is in Max SNP if it can be expressed as opt Q (A) = max {x φ(x, S, A)}. S Read: satisfy the (constant-length) formula φ(x, S, A) for as many tuples x as possible Example (Max Cut): max S V {(u, v) (E(u, v) S(u) S(v))} Max SNP APX [Papadimitriou & Yannakakis 91]

23 12 Polynomial kernel for Max SNP given (A, k) decide whether max S {x φ(x, S, A)} k a random choice of S satisfies φ(x, S, A) for a constant fraction 1 c of the tuples x for which it is satisfiable if there are more than ck tuples then report YES else there are few satisfiable tuples by shrinking A we get a kernel

24 13 A problem Q is in if it can be expressed as opt Q (A) = max {x ( y) : φ(x, y, S, A)}. S Read: satisfy the long formula ( y) : φ(x, y, S, A) for as many tuples x as possible Example (Max Sat): max T {c ( v) : (P(v, c) T (v)) (N(v, c) T (v))} APX [Papadimitriou & Yannakakis 91]

25 14 Polynomial kernel for given (A, k) decide whether opt Q (A) k again, we are done if there are ck satisfiable tuples, else there are less than ck however, the formulas ( y) : φ(x, y, S, A) may be long φ(x, y, S, A) has constant size and a constant number s of occurrences of S (= literals) satisfying one φ(x, y, S, A) for some y suffices for x assignment to sk variables suffices for k tuples x use the Sunflower Lemma to reduce the number of y s shrink A to get a polynomial kernel

26 Polynomial kernel for Example: φ(x, y 1, S, A),..., φ(x, y 6, S, A) share exactly the literals L 0 L 3 L 2 L 0 L 1 L 4 L 5 L 6 assignment to four variables contradicts L0 or at most four sets L i may discard one formula 15

27 16 Min F + Π 1 A problem Q is in Min F + Π 1 if it can be expressed as opt Q (A) = min S { S ( x) : φ(x, S, A)}, where φ is in CNF and positive in S. Read: Satisfy ( x) : φ(x, S, A) with as few tuples in S as possible (note, S(x) = true x S). Example (Vertex Cover): min S V { S ( (u, v) V 2 ) : ( E(u, v) S(u) S(v)) } Min F + Π 1 APX [Kolaitis & Thakur 94, 95]

28 17 Polynomial kernel for Min F + Π 1 for each input A, ( x) : φ(x, S, A) encodes a d-hitting Set instance (H(A), k): each tuple x gives rise to a formula φ(x, S, A) each formula gives some sets, since tuples must be added to S to satisfy it use a d-hs kernelization that only deletes sets H keep only tuples x that contribute sets in H and shrink A to get a polynomial kernel

29 18 Outline

30 19 a Boolean relation: R {0, 1} r a constraint: R(x 1,..., x r ) a constraint language: finite set Γ of Boolean relations a formula over Γ: R 1 (x 11,..., x 1r1 )... R t (x t1,..., x trt ) (i) each R i is some relation from Γ (ii) the x ij are Boolean variables

31 20 Problem: Given F over Γ find a (satisfying) assignment that optimizes a certain goal. possible goals: 1) maximize # satisfied constraints 2) minimize # unsatisfied constraints 3) maximize/minimize # true variables in a satisfying assignment parameterized version: opt k? (opt k?) with parameter k

32 21 Some types of constraints 0-valid (0,..., 0) R 1-valid (1,..., 1) R affine x 1... x t = d, for d {0, 1} IHS-B positive clauses, assignments and implications, or negative clauses,... bijunctive can be expressed by 2CNF formula Horn can be expressed by CNF formula with at most one positive literal per clause dual-horn... at most one negative literal per clause 2-monotone can be expressed by DNF formula with at most one positive and one negative disjunct/term

33 22 Max SAT(Γ) Max SAT(Γ) Input: A formula F over Γ. Output: An assignment that satisfies a maximum number of constraints of F. Example: Choosing Γ = {(x y)} gives Max Cut Max SAT(Γ)......is in P for 0-valid, 1-valid, and 2-monotone Γ...is in APX for all Γ...admits a polynomial kernel for all Γ

34 23 Min UnSAT(Γ) Min UnSAT(Γ) Input: A formula F over Γ. Output: An assignment that dissatisfies a minimum number of constraints of F. Min UnSAT(Γ)......is in P for 0-valid, 1-valid, and 2-monotone Γ....is in APX when Γ is IHS-B...has open parameterized complexity, except for: Almost 2SAT is FPT [O Sullivan & Razgon 08] NP-hard for k=0 when SAT(Γ) is NP-hard

35 24 Max Ones SAT(Γ) Max Ones SAT(Γ) Input: A formula F over Γ. Output: A satisfying assignment with a maximum number of true variables. Example: Choosing Γ = {( x y)} gives Independent Set Max Ones SAT(Γ)......is in P for 1-valid, dual-horn, and width-2 affine Γ...is in APX for affine Γ...admits a polynomial kernel for affine Γ (complete classification in [K. et al. 10])

36 25 Polynomial kernel for Max Ones SAT(Γ) Let (F, k) be an instance of Max Ones SAT(Γ) for some affine Γ: if F is unsatisfiable return NO

37 25 Polynomial kernel for Max Ones SAT(Γ) Let (F, k) be an instance of Max Ones SAT(Γ) for some affine Γ: if F is unsatisfiable return NO Check for fixed variables: if F (x = 0) is unsatisfiable x must be assigned 1 if F (x = 1) is unsatisfiable x must be assigned 0

38 25 Polynomial kernel for Max Ones SAT(Γ) Let (F, k) be an instance of Max Ones SAT(Γ) for some affine Γ: if F is unsatisfiable return NO Check for fixed variables: if F (x = 0) is unsatisfiable x must be assigned 1 if F (x = 1) is unsatisfiable x must be assigned 0 we will show that at least half of the other variables can be assigned true

39 26 Polynomial kernel for Max Ones SAT(Γ) At least half of the unfixed variables can be set to 1: pick an unfixed variable x

40 26 Polynomial kernel for Max Ones SAT(Γ) At least half of the unfixed variables can be set to 1: pick an unfixed variable x Y0 : additional fixed variables in F (x = 0) Y1 : additional fixed variables in F (x = 1) Y0 = Y 1 since Γ is affine

41 26 Polynomial kernel for Max Ones SAT(Γ) At least half of the unfixed variables can be set to 1: pick an unfixed variable x Y0 : additional fixed variables in F (x = 0) Y1 : additional fixed variables in F (x = 1) Y0 = Y 1 since Γ is affine if x = 0 fixes y to 0 then x = 1 fixes y to 1; and vice versa make the choice for x that fixes more variables to 1 repeat

42 27 How to get the kernel for Max Ones SAT(Γ) identify all fixed to 0 variables to a single variable identify all fixed to 1 variables to a single variable; update k

43 27 How to get the kernel for Max Ones SAT(Γ) identify all fixed to 0 variables to a single variable identify all fixed to 1 variables to a single variable; update k if 2k unfixed variables return YES else return kernel with < 2k + 2 variables

44 28 Min Ones SAT(Γ) Min Ones SAT(Γ) Input: A formula F over Γ. Output: A satisfying assignment with a minimum number of true variables. Min Ones SAT(Γ)......is in P for 0-valid, Horn, and width-2 affine Γ...is in APX when Γ is IHS-B or bijunctive...admits a polynomial kernel when Γ is bijunctive or IHS-B (complete classification in [K. & Wahlström 10])

45 29 Polynomial kernel for Min Ones SAT(Γ) We show: Polynomial kernel for Min Ones SAT(Γ) when Γ is IHS-B or bijunctive. It suffices to consider the following d-hitting Set variant: a family of sets H each of size d for which a hitting set of size k is requested a second family of sets F; none of its sets may be subsets of the hitting set a set of implications, i.e., if x is selected for the hitting set, then y must be selected too

46 30 Polynomial kernel for Min Ones SAT(Γ) delete all elements whose selection implies > k selections

47 30 Polynomial kernel for Min Ones SAT(Γ) delete all elements whose selection implies > k selections when an element is deleted: remove it from all sets of H delete all sets of F that contain it delete all elements that would imply it

48 30 Polynomial kernel for Min Ones SAT(Γ) delete all elements whose selection implies > k selections when an element is deleted: remove it from all sets of H delete all sets of F that contain it delete all elements that would imply it use sunflower-based kernelization to shrink the set family H to H of size O(k d )

49 30 Polynomial kernel for Min Ones SAT(Γ) delete all elements whose selection implies > k selections when an element is deleted: remove it from all sets of H delete all sets of F that contain it delete all elements that would imply it use sunflower-based kernelization to shrink the set family H to H of size O(k d ) delete all elements that are not in a set of H or implied by such an element kernel with O(k d+1 ) elements

50 31 Outline

51 32 Bin Bin Input: A set of n items s 1,..., s n [0, 1]; an integer k. Parameter: k. Output: Report whether the n items can be packed into k bins of size 1.

52 32 Bin Bin Input: A set of n items s 1,..., s n [0, 1]; an integer k. Parameter: k. Output: Report whether the n items can be packed into k bins of size 1. Partition Input: A set of n integers. Output: Decide whether the set can be partitioned into two sets of equal sum.

53 32 Bin Bin Input: A set of n items s 1,..., s n [0, 1]; an integer k. Parameter: k. Output: Report whether the n items can be packed into k bins of size 1. Partition Input: A set of n integers. Output: Decide whether the set can be partitioned into two sets of equal sum. NP-hard for k = 2 by a reduction from Partition not in XP unless P = NP

54 33 Input: A set of n items s 1,..., s n [0, 1]; an integer k. Parameter: k. Output: A packing into k + 1 bins each of size one or report that k bins do not suffice.

55 33 Input: A set of n items s 1,..., s n [0, 1]; an integer k. Parameter: k. Output: A packing into k + 1 bins each of size one or report that k bins do not suffice. Note: 2 O(k log2 k) n O(1) time algorithm [Jansen et al. 2010]

56 34 A reduction to O(k 2 ) items Given a set of n items s 1,..., s n [0, 1] and an integer k:

57 34 A reduction to O(k 2 ) items Given a set of n items s 1,..., s n [0, 1] and an integer k: if total weight si > k k bins do not suffice

58 A reduction to O(k 2 ) items Given a set of n items s 1,..., s n [0, 1] and an integer k: if total weight si > k k bins do not suffice forget all items of size less than 1 k+1 34

59 A reduction to O(k 2 ) items Given a set of n items s 1,..., s n [0, 1] and an integer k: if total weight si > k k bins do not suffice forget all items of size less than 1 k+1 at most k(k + 1) items have size 1 k+1 34

60 35 Correctness Let S be obtained from S = {s 1,..., s n } by deleting items smaller than 1 k+1 and assume s i k.

61 Correctness Let S be obtained from S = {s 1,..., s n } by deleting items smaller than 1 k+1 and assume s i k. if S can be packed into k + 1 bins then so can S 35

62 Correctness Let S be obtained from S = {s 1,..., s n } by deleting items smaller than 1 k+1 and assume s i k. if S can be packed into k + 1 bins then so can S consider any packing for S and greedily add the small items from S \ S : 35

63 Correctness Let S be obtained from S = {s 1,..., s n } by deleting items smaller than 1 k+1 and assume s i k. if S can be packed into k + 1 bins then so can S consider any packing for S and greedily add the small items from S \ S : if all items fit then done 35

64 Correctness Let S be obtained from S = {s 1,..., s n } by deleting items smaller than 1 k+1 and assume s i k. if S can be packed into k + 1 bins then so can S consider any packing for S and greedily add the small items from S \ S : if all items fit then done else the free space in each of the bins is less than 1 k+1 35

65 35 Correctness Let S be obtained from S = {s 1,..., s n } by deleting items smaller than 1 k+1 and assume s i k. if S can be packed into k + 1 bins then so can S consider any packing for S and greedily add the small items from S \ S : if all items fit then done else the free space in each of the bins is less than 1 k+1 thus total weight in the bins would be greater than ( (k + 1) 1 1 ) = k k + 1

66 35 Correctness Let S be obtained from S = {s 1,..., s n } by deleting items smaller than 1 k+1 and assume s i k. if S can be packed into k + 1 bins then so can S consider any packing for S and greedily add the small items from S \ S : if all items fit then done else the free space in each of the bins is less than 1 k+1 thus total weight in the bins would be greater than ( (k + 1) 1 1 ) = k k + 1 S can be packed into k + 1 bins

67 36 Outline

68 37 noticable trend: APX PK not true in general (Bin, Connected Vertex Cover, Unit Disc IS) holds when restricted to certain CSPs or syntactical problems FPTAS LK by a simple variation on a folklore proof (and eff-fpt LK) even for Bin there is some good news

69 38 Thank you