Minimum Bisection is fixed-parameter tractable

Transcription

1 Minimum Bisection is fixed-parameter tractable Marek Cygan 1, Daniel Lokshtanov 2, Marcin Pilipczuk 2, Micha l Pilipczuk 2, Saket Saurabh 2,3 1 Institute of Informatics, University of Warsaw, Poland 2 Department of Informatics, University of Bergen, Norway 3 Institute of Mathematical Sciences, Chennai, India Bertinoro Workshop on Graphs, December 17 th, 2013 Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 1/16

2 The problem Mininimum Bisection Input: Question: A graph G with V (G) = 2n and an integer k Does there exist a partition (A, B) of V (G) such that A = B = n and E(A, B) k? Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 2/16

3 Example Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 3/16

4 Example Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 3/16

5 About the problem MinBisection is one of the most classic NP-hard problems, Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 4/16

6 About the problem MinBisection is one of the most classic NP-hard problems, but also one of the most mysterious ones. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 4/16

7 About the problem MinBisection is one of the most classic NP-hard problems, but also one of the most mysterious ones. Only O(log n)-approximation is known [Räcke], but a constant factor approximation is not refuted. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 4/16

8 About the problem MinBisection is one of the most classic NP-hard problems, but also one of the most mysterious ones. Only O(log n)-approximation is known [Räcke], but a constant factor approximation is not refuted. It is open whether MinBisection is NP-hard on planar graphs. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 4/16

9 About the problem MinBisection is one of the most classic NP-hard problems, but also one of the most mysterious ones. Only O(log n)-approximation is known [Räcke], but a constant factor approximation is not refuted. It is open whether MinBisection is NP-hard on planar graphs. Our result: MinBisection is fixed-parameter tractable when parameterized by k. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 4/16

10 About the problem MinBisection is one of the most classic NP-hard problems, but also one of the most mysterious ones. Only O(log n)-approximation is known [Räcke], but a constant factor approximation is not refuted. It is open whether MinBisection is NP-hard on planar graphs. Our result: MinBisection is fixed-parameter tractable when parameterized by k. Running time: 2 O(k3) n 3 log 3 n Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 4/16

11 The approach Decomposition: a new decomposition theorem for graphs Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 5/16

12 The approach Decomposition: a new decomposition theorem for graphs Intuition: A graph can be decomposed in a tree-like manner by separators of size roughly k into pieces that are highly connected in terms of k, and cannot be broken further. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 5/16

13 The approach Decomposition: a new decomposition theorem for graphs Intuition: A graph can be decomposed in a tree-like manner by separators of size roughly k into pieces that are highly connected in terms of k, and cannot be broken further. More formally, each bag of the tree decomposition is (q, k)-unbreakable for q = f (k). Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 5/16

14 The approach Decomposition: a new decomposition theorem for graphs Intuition: A graph can be decomposed in a tree-like manner by separators of size roughly k into pieces that are highly connected in terms of k, and cannot be broken further. More formally, each bag of the tree decomposition is (q, k)-unbreakable for q = f (k). Dynamic programming: run a bottom-up DP on the obtained tree decomposition. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 5/16

15 The approach Decomposition: a new decomposition theorem for graphs Intuition: A graph can be decomposed in a tree-like manner by separators of size roughly k into pieces that are highly connected in terms of k, and cannot be broken further. More formally, each bag of the tree decomposition is (q, k)-unbreakable for q = f (k). Dynamic programming: run a bottom-up DP on the obtained tree decomposition. MinBisection can be robustly solved in FPT time on (q, k)-unbreakable graphs using randomized contractions. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 5/16

16 The approach Decomposition: a new decomposition theorem for graphs Intuition: A graph can be decomposed in a tree-like manner by separators of size roughly k into pieces that are highly connected in terms of k, and cannot be broken further. More formally, each bag of the tree decomposition is (q, k)-unbreakable for q = f (k). Dynamic programming: run a bottom-up DP on the obtained tree decomposition. MinBisection can be robustly solved in FPT time on (q, k)-unbreakable graphs using randomized contractions. Use this algorithm as a subroutine at each bag of the decomposition. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 5/16

17 The approach Decomposition: a new decomposition theorem for graphs Intuition: A graph can be decomposed in a tree-like manner by separators of size roughly k into pieces that are highly connected in terms of k, and cannot be broken further. More formally, each bag of the tree decomposition is (q, k)-unbreakable for q = f (k). Dynamic programming: run a bottom-up DP on the obtained tree decomposition. MinBisection can be robustly solved in FPT time on (q, k)-unbreakable graphs using randomized contractions. Use this algorithm as a subroutine at each bag of the decomposition. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 5/16

18 Unbreakable sets (A, B) is a separation if A B = V (G) and E(A \ B, B \ A) =. A B is the separator, and A B is the order of the separation. A B A B Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 6/16

19 Unbreakable sets (A, B) is a separation if A B = V (G) and E(A \ B, B \ A) =. A B is the separator, and A B is the order of the separation. (q, k)-unbreakability We say that a subset X V (G) is (q, k)-unbreakable if for every separation (A, B) of G of order at most k, either (A \ B) X q or (B \ A) X q. A B A B Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 6/16 X

20 Unbreakable sets (A, B) is a separation if A B = V (G) and E(A \ B, B \ A) =. A B is the separator, and A B is the order of the separation. (q, k)-unbreakability We say that a subset X V (G) is (q, k)-unbreakable if for every separation (A, B) of G of order at most k, either (A \ B) X q or (B \ A) X q. Intuition: a separation of small order can carve out only a small portion of X. A B A B Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 6/16 X

21 The decomposition theorem The decomposition theorem Let G be a graph and k be an integer. Then there exists a tree decomposition T = (T, {B u } u V (T ) ) of G such that: Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 7/16

22 The decomposition theorem The decomposition theorem Let G be a graph and k be an integer. Then there exists a tree decomposition T = (T, {B u } u V (T ) ) of G such that: Each B u is (τ, k)-unbreakable in G, where τ = 2 O(k). Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 7/16

23 The decomposition theorem The decomposition theorem Let G be a graph and k be an integer. Then there exists a tree decomposition T = (T, {B u } u V (T ) ) of G such that: Each B u is (τ, k)-unbreakable in G, where τ = 2 O(k). B u B v η for each uv E(T ), where η = 2 O(k). Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 7/16

24 The decomposition theorem The decomposition theorem Let G be a graph and k be an integer. Then there exists a tree decomposition T = (T, {B u } u V (T ) ) of G such that: Each B u is (τ, k)-unbreakable in G, where τ = 2 O(k). B u B v η for each uv E(T ), where η = 2 O(k). B u B v is (2k, k)-unbreakable in G, for each uv E(T ). Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 7/16

25 The decomposition theorem The decomposition theorem Let G be a graph and k be an integer. Then there exists a tree decomposition T = (T, {B u } u V (T ) ) of G such that: Each B u is (τ, k)-unbreakable in G, where τ = 2 O(k). B u B v η for each uv E(T ), where η = 2 O(k). B u B v is (2k, k)-unbreakable in G, for each uv E(T ). Moreover, this decomposition can be computed in 2 O(k2) n 2 m time. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 7/16

26 On a picture Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 8/16

27 On a picture (τ, k)-unbreakable Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 8/16

28 On a picture η, and (2k, k)-unbreakable Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 8/16

29 A glimpse into the proof Now: A sketch of the proof, without unbreakability of adhesions. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 9/16

30 A glimpse into the proof Now: A sketch of the proof, without unbreakability of adhesions. Construct the decomposition in a standard top-down manner. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 9/16

31 A glimpse into the proof Now: A sketch of the proof, without unbreakability of adhesions. Construct the decomposition in a standard top-down manner. Problem to solve at one step: Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 9/16

32 A glimpse into the proof Now: A sketch of the proof, without unbreakability of adhesions. Construct the decomposition in a standard top-down manner. Problem to solve at one step: Input: Subgraph H G to be decomposed, and a set S V (H), S η, that will be the top adhesion. S H Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 9/16

33 A glimpse into the proof Now: A sketch of the proof, without unbreakability of adhesions. Construct the decomposition in a standard top-down manner. Problem to solve at one step: Input: Subgraph H G to be decomposed, and a set S V (H), S η, that will be the top adhesion. Output: A candidate unbreakable bag A S, such that every connected component of H \ A sees only η vertices in A. S A Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 9/16

34 A glimpse into the proof Now: A sketch of the proof, without unbreakability of adhesions. Construct the decomposition in a standard top-down manner. Problem to solve at one step: Input: Subgraph H G to be decomposed, and a set S V (H), S η, that will be the top adhesion. Output: A candidate unbreakable bag A S, such that every connected component of H \ A sees only η vertices in A. Then recurse into the connected components below. S A Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 9/16

35 Breakable S Test whether S is (k, k)-breakable in H. A Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 10/16

36 Breakable S Test whether S is (k, k)-breakable in H. If so, then just break. k A k k Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 10/16

37 Breakable S Test whether S is (k, k)-breakable in H. If so, then just break. A η + k, so A is (τ, k)-unbreakable for any τ η + k. k A k k Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 10/16

38 Breakable S Test whether S is (k, k)-breakable in H. If so, then just break. A η + k, so A is (τ, k)-unbreakable for any τ η + k. Every connected component below is either on the left or on the right, so it sees only η vertices from A. k A k k Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 10/16

39 Unbreakable S Assume then that S is (k, k)-unbreakable. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 11/16

40 Unbreakable S Assume then that S is (k, k)-unbreakable. Every separation (A, B) of order k in H has a small side (with at most k vertices from S), and a large size (with the rest). S Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 11/16

41 Unbreakable S Assume then that S is (k, k)-unbreakable. Every separation (A, B) of order k in H has a small side (with at most k vertices from S), and a large size (with the rest). Adding these k vertices to the large side gives a separation of order 2k that carves out a bunch of vertices from S. S 2k Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 11/16

42 Unbreakable S Assume then that S is (k, k)-unbreakable. Every separation (A, B) of order k in H has a small side (with at most k vertices from S), and a large size (with the rest). Adding these k vertices to the large side gives a separation of order 2k that carves out a bunch of vertices from S. Idea: Mark greedily all possible such carvings; what is left constitutes A. S 2k Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 11/16

43 Important separators u S Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 12/16

44 Important separators u X S X is an important u-s separator if Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 12/16

45 Important separators u X S X is an important u-s separator if X is a minimal u-s separator, and Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 12/16

46 Important separators u X X S X is an important u-s separator if X is a minimal u-s separator, and there is no u-s separator X such that (i) X X and (ii) reach(u, H \ X ) reach(u, H \ X ). Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 12/16

47 Important separators u X S X is an important u-s separator if X is a minimal u-s separator, and there is no u-s separator X such that (i) X X and (ii) reach(u, H \ X ) reach(u, H \ X ). Note: minimality X = N(reach(u, H \ X )). Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 12/16

48 Important separators u X S X is an important u-s separator if X is a minimal u-s separator, and there is no u-s separator X such that (i) X X and (ii) reach(u, H \ X ) reach(u, H \ X ). Note: minimality X = N(reach(u, H \ X )). For fixed u and S, there are at most 4 k important u-s separators of size at most k. [Chen et al., Marx and Razgon] Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 12/16

49 Chips and the decomposition S H Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 13/16

50 Chips and the decomposition S For every u V (H), list all the important separators u-s of size at most 2k. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 13/16

51 Chips and the decomposition S For every u V (H), list all the important separators u-s of size at most 2k. For each separator, take the set of vertices reachable from u. Filter out all such sets that are not inclusion-wise maximal. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 13/16

52 Chips and the decomposition S For every u V (H), list all the important separators u-s of size at most 2k. For each separator, take the set of vertices reachable from u. Filter out all such sets that are not inclusion-wise maximal. The obtained sets will be called chips, and the family of chips is called C. Note that C 4 2k n. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 13/16

53 Chips and the decomposition S For every u V (H), list all the important separators u-s of size at most 2k. For each separator, take the set of vertices reachable from u. Filter out all such sets that are not inclusion-wise maximal. The obtained sets will be called chips, and the family of chips is called C. Note that C 4 2k n. Set A = C C N(C) C C V (H) \ N[C]. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 13/16

54 Bounding the adhesions Examine one connected component D of H \ A and suppose that it is contained in chip C. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 14/16

55 Bounding the adhesions Examine one connected component D of H \ A and suppose that it is contained in chip C. N(D) is contained in N(C) plus the neighbourhoods of all chips C that interfere with C. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 14/16

56 Bounding the adhesions C C Examine one connected component D of H \ A and suppose that it is contained in chip C. N(D) is contained in N(C) plus the neighbourhoods of all chips C that interfere with C. C and C touch if C C or E(C, C ). Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 14/16

57 Bounding the adhesions C C u Examine one connected component D of H \ A and suppose that it is contained in chip C. N(D) is contained in N(C) plus the neighbourhoods of all chips C that interfere with C. C and C touch if C C or E(C, C ). Connectivity There exists u N(C) C. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 14/16

58 Bounding the adhesions, continued S C C u Observation: N(C ) is an important u-s separator. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 15/16

59 Bounding the adhesions, continued S C C u Observation: N(C ) is an important u-s separator. Hence, every chip touching C is raised by an important separator from a vertex of N(C). Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 15/16

60 Bounding the adhesions, continued S C C u Observation: N(C ) is an important u-s separator. Hence, every chip touching C is raised by an important separator from a vertex of N(C). Ergo every chip touches at most 2k 4 2k other chips, and every component of H \ A sees at most (2k) (2k + 1) 4 2k vertices. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 15/16

61 Bounding the adhesions, continued S Observation: N(C ) is an important u-s separator. Hence, every chip touching C is raised by an important separator from a vertex of N(C). Ergo every chip touches at most 2k 4 2k other chips, and every component of H \ A sees at most (2k) (2k + 1) 4 2k vertices. Check: A is (τ, k)-unbreakable in H, for some τ = 2 O(k). Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 15/16

62 Open problems Many open questions: Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 16/16

63 Open problems Many open questions: Are exponential-size adhesions necessary? Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 16/16

64 Open problems Many open questions: Are exponential-size adhesions necessary? Does unbreakability need to depend exponentially on k? Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 16/16

65 Open problems Many open questions: Are exponential-size adhesions necessary? Does unbreakability need to depend exponentially on k? Can we compute the decomposition quicker? Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 16/16

66 Open problems Many open questions: Are exponential-size adhesions necessary? Does unbreakability need to depend exponentially on k? Can we compute the decomposition quicker? Both terms 2 O(k2) and n 2 m seem suboptimal. Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 16/16

67 Open problems Many open questions: Are exponential-size adhesions necessary? Does unbreakability need to depend exponentially on k? Can we compute the decomposition quicker? Both terms 2 O(k2) and n 2 m seem suboptimal. Can we find other applications of the decomposition theorem? Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 16/16

68 Open problems Many open questions: Are exponential-size adhesions necessary? Does unbreakability need to depend exponentially on k? Can we compute the decomposition quicker? Both terms 2 O(k2) and n 2 m seem suboptimal. Can we find other applications of the decomposition theorem? Thanks for attention! Cygan, Lokshtanov, Pilipczuk 2, Saurabh Minimum Bisection is FPT 16/16