EPTAS for Maximum Clique on Disks and Unit Balls

Transcription

1 EPTAS for Maximum Clique on Disks and Unit Balls Édouard Bonnet joint work with Panos Giannopoulos, Eunjung Kim, Paweł Rzążewski, and Florian Sikora and Marthe Bonamy, Nicolas Bousquet, Pierre Chabit, and Stéphan Thomassé LIP, ENS Lyon May 18th 2018, Leiden

2 Find a largest collection of disks that pairwise intersect

3 Like this

4 or that

5 Polynomial algorithm on unit disks by Clark, Colbourn, and Johnson, c i c j Guess two farthest disks in an optimum solution S.

6 Polynomial algorithm on unit disks by Clark, Colbourn, and Johnson, c i c j Hence, all the centers of S lie inside the bold digon.

7 Polynomial algorithm on unit disks by Clark, Colbourn, and Johnson, c i c j Two disks centered in the same-color region intersect.

8 Polynomial algorithm on unit disks by Clark, Colbourn, and Johnson, c i c j We solve Max Clique in a co-bipartite graph.

9 Polynomial algorithm on unit disks by Clark, Colbourn, and Johnson, c i c j We solve Max Independent Set in a bipartite graph.

10 Disk graphs Inherits the NP-hardness of planar graphs.

11 So what is known for Max Clique on disk graphs? Polynomial-time 2-approximation For any clique there are 4 points hitting all the disks. Guess those points and remove the non-hit disks. The resulting graph is partitioned into 2 co-bipartite graphs. Solve exactly on both co-bipartite graphs. Output the best solution. No non-trivial exact algorithm known

12 And what is known about disk graphs? Every planar graph is a disk graph. Every triangle-free disk graph is planar (centers vertices). So a triangle-free non-planar graph like K 3,3 is not disk. A subdivision of a non-planar graph is not a disk graph (more generally not a string graph)....

13 And what is known about disk graphs? Every planar graph is a disk graph. Every triangle-free disk graph is planar (centers vertices). So a triangle-free non-planar graph like K 3,3 is not disk. A subdivision of a non-planar graph is not a disk graph (more generally not a string graph).... Other ways of showing that a graph is not disk?

14 Say the 4 centers encoding a K 2,2 = 2K 2 are in convex position. c 3 c 2 c 1 c 4

15 Say the 4 centers encoding a K 2,2 = 2K 2 are in convex position. c 3 c 2 c 1 c 4 Then the two non-edges should be diagonal.

16 Say the 4 centers encoding a K 2,2 = 2K 2 are in convex position. c 3 c 2 c 1 c 4 Then the two non-edges should be diagonal. Suppose d(c 1, c 3 ) > r 1 + r 3 and d(c 2, c 4 ) > r 2 + r 4. But d(c 1, c 3 ) + d(c 2, c 4 ) d(c 1, c 2 ) + d(c 3, c 4 ) r 1 + r 2 + r 3 + r 4, a contradiction.

17 Conclusion: the 4 centers of an induced 2K 2 are either not in convex position or in convex position with the non-edges being diagonal. c 2 c 3 c4 or c 1 c 3 c 1 c 2 c 4

18 Conclusion: the 4 centers of an induced 2K 2 are either not in convex position or in convex position with the non-edges being diagonal. c 2 c 3 c4 or c 1 c 3 c 1 c 2 c 4 Reformulation: either the line l(c 1, c 2 ) crosses the segment c 3 c 4, or the line l(c 3, c 4 ) crosses the segment c 1 c 2, or both; equivalently, the segments c 1 c 2 and c 3 c 4 cross.

19 Assume C s + C t is a disk graph. Link consecutive centers of the two disjoint cycles (non-edges). For each red segment s i, we denote by: a i the number of blue segments crossed by l(s i ). b i the number of blue segments whose extension cross s i. c i the number of blue segments intersecting s i.

20 Assume C s + C t is a disk graph. Link consecutive centers of the two disjoint cycles (non-edges). s i For each red segment s i, we denote by: a i the number of blue segments crossed by l(s i ). b i the number of blue segments whose extension cross s i. c i the number of blue segments intersecting s i.

21 Assume C s + C t is a disk graph. Link consecutive centers of the two disjoint cycles (non-edges). s i For each red segment s i, we denote by: a i the number of blue segments crossed by l(s i ). b i the number of blue segments whose extension cross s i. c i the number of blue segments intersecting s i.

22 Assume C s + C t is a disk graph. Link consecutive centers of the two disjoint cycles (non-edges). s i For each red segment s i, we denote by: a i the number of blue segments crossed by l(s i ). b i the number of blue segments whose extension cross s i. c i the number of blue segments intersecting s i.

23 Assume C s + C t is a disk graph. Link consecutive centers of the two disjoint cycles (non-edges). s i For each red segment s i, we denote by: a i the number of blue segments crossed by l(s i ). b i the number of blue segments whose extension cross s i. c i the number of blue segments intersecting s i. It should be that a i + b i c i = t.

24 Σ a i + b i c i = st 1 i s 1) a i is even:

25 Σ a i + b i c i = st 1 i s 1) a i is even: number of intersections of a line with a closed curve. 2) Σ 1 i s b i =

26 Σ a i + b i c i = st 1 i s 1) a i is even: number of intersections of a line with a closed curve. 2) Σ b i = Σ a i is therefore even. (a j, b j, c j same for blue segments) 1 i s 1 i t 3) Σ 1 i s c i is even:

27 Σ a i + b i c i = st 1 i s 1) a i is even: number of intersections of a line with a closed curve. 2) Σ b i = Σ a i is therefore even. (a j, b j, c j same for blue segments) 1 i s 1 i t 3) Σ 1 i s c i is even: number of intersections of two closed curves. Σ a i + b i c i = 1 i s

28 Σ a i + b i c i = st 1 i s 1) a i is even: number of intersections of a line with a closed curve. 2) Σ b i = Σ a i is therefore even. (a j, b j, c j same for blue segments) 1 i s 1 i t 3) Σ 1 i s c i is even: number of intersections of two closed curves. Σ a i + b i c i = 1 i s Σ a i + Σ a i Σ c i is even. 1 i s 1 i t 1 i s

29 Σ a i + b i c i = st 1 i s 1) a i is even: number of intersections of a line with a closed curve. 2) Σ b i = Σ a i is therefore even. (a j, b j, c j same for blue segments) 1 i s 1 i t 3) Σ 1 i s c i is even: number of intersections of two closed curves. Σ a i + b i c i = 1 i s Σ a i + Σ a i Σ c i is even. 1 i s 1 i t 1 i s Hence s and t cannot be both odd.

30 The complement of two odd cycles is not a disk graph. Are there other graphs of co-degree 2 which are not disk?

31 Complement of many even cycles and one odd cycle D 2i+1 D 2i+1 D 1 D 2s+1 D 2i

32 Can we solve Max Independent Set more efficiently if there are no two vertex-disjoint odd cycles as an induced subgraph?

33 Can we solve Max Independent Set more efficiently if there are no two vertex-disjoint odd cycles as an induced subgraph? Another way to see it: at least one edge between two vertex-disjoint odd cycles

34 Can we solve Max Independent Set more efficiently if there are no two vertex-disjoint odd cycles as an induced subgraph? Another way to see it: at least one edge between two vertex-disjoint odd cycles We can get a QPTAS and an exact subexponential algorithm in 2Õ(n2/3) with win-wins and known results.

35 EPTAS VC dim of S = maximum size of a set with all intersections with S. VCdim(G) = VC dimension of the neighborhood set-system. α(g) = size of a maximum independent set in G. iocp(g) = same as ocp but induced.

36 EPTAS VC dim of S = maximum size of a set with all intersections with S. VCdim(G) = VC dimension of the neighborhood set-system. α(g) = size of a maximum independent set in G. iocp(g) = same as ocp but induced. Theorem Max Independent Set can be 1 + ε-approximated in time 2Õ(1/ε3) n O(1) on graphs G with VCdim(G) = O(1), α(g) = Ω( V (G) ), and iocp(g) = 1.

37 Classic result of Haussler and Welzl in VC dimension theory Theorem (ε-nets) A set-system (S, U) with VC dimension d and only sets of size at least ε U has a hitting set of size O( d ε log 1 ε ). Furthermore, any sample of size 10d ε log 1 ε is a hitting set w.h.p.

38 Classic result of Haussler and Welzl in VC dimension theory Theorem (ε-nets) A set-system (S, U) with VC dimension d and only sets of size at least ε U has a hitting set of size O( d ε log 1 ε ). Furthermore, any sample of size 10d ε log 1 ε is a hitting set w.h.p. We will apply that result to the set-system ({N(u) I u V (G), N(u) I ε 3 I }, I). In words, the large neighborhoods over I.

39 First step: sampling I is a fixed maximum independent set. We can assume that I = Θ(n). G I We pick randomly S of Õ(1/ε 3 ) vertices.

40 First step: sampling I is a fixed maximum independent set. We can assume that I = Θ(n). G I S With probability f (ε) > 0, S I.

41 First step: sampling I is a fixed maximum independent set. We can assume that I = Θ(n). G N(S) I S We delete the neighborhood of S.

42 First step: sampling I is a fixed maximum independent set. We can assume that I = Θ(n). G N(S) I S < ε 3 I The remaining vertices have few vertices in I.

43 First step: sampling I is a fixed maximum independent set. We can assume that I = Θ(n). G N(S) I S < ε 3 I This is due to the theorem of ε-nets.

44 First step: sampling I is a fixed maximum independent set. We can assume that I = Θ(n). G N(S) I S We compute a shortest odd cycle C.

45 First step: sampling I is a fixed maximum independent set. We can assume that I = Θ(n). G N(S) I S < ε I If C 1/ε 2, we delete its neighborhood.

46 First step: sampling I is a fixed maximum independent set. We can assume that I = Θ(n). G N(S) I S So, we might assume that C > 1/ε 2.

47 Second step: find a small odd cycle transversal In column, the successive neighborhood of C, layers. Rows indicate the closest neighbor on C, strata.

48 Second step: find a small odd cycle transversal bipartite Deleting a j-th neighborhood of C, leaves a bipartite to the right.

49 Second step: find a small odd cycle transversal bipartite Deleting a j-th neighborhood of C, leaves a bipartite to the right.

50 Second step: find a small odd cycle transversal bipartite Deleting a j-th neighborhood of C, leaves a bipartite to the right.

51 Second step: find a small odd cycle transversal bipartite We delete the lightest of the 1/ε first layers.

52 Second step: find a small odd cycle transversal bipartite bipartite The 1/ε consecutive layers form an odd cycle transversal.

53 Second step: find a small odd cycle transversal bipartite bipartite The 1/ε consecutive layers form an odd cycle transversal.

54 Second step: find a small odd cycle transversal bipartite bipartite The 1/ε consecutive layers form an odd cycle transversal.

55 Second step: find a small odd cycle transversal bipartite bipartite We remove the lightest block of strata.

56 Filled ellipses and triangles 2-subdivisions: graphs where each edge is subdivided exactly twice co-2-subdivisions: complements of 2-subdivisions Lemma For some α > 1, Max Independent Set on 2-subdivisions is not α-approximable algorithm in 2 n0.99, unless the ETH fails.

57 Filled ellipses and triangles 2-subdivisions: graphs where each edge is subdivided exactly twice co-2-subdivisions: complements of 2-subdivisions Lemma For some α > 1, Max Independent Set on 2-subdivisions is not α-approximable algorithm in 2 n0.99, unless the ETH fails. Graphs of filled ellipses or filled triangles contain all the co-2-subdivisions.

58 Filled triangles

59 Filled ellipses

60 Higher dimensions: unit 4D-disk graphs ball (3D-disk) graphs with radii arbitrary close to 1 contain all the co-2-subdivisions (so, no approximation scheme and no subexponential algorithm) p(v 1 ) P 3 ε P π(p + (e 1 )) O 1 C π(p (e 1 ))

61 What about unit ball graph? Let x 1,..., x s be the consecutive centers in R 3 of a co-odd-cycle. Consider the trace on the 2-sphere of the following vector walk. Start at vector ab with a = x 1 and b = x 2. move continuously a from x 1 to x 3 following the segment x 1 x 3. move continuously b from x 2 to x 4 following the segment x 2 x 4. and so on, until back to x 1 x 2.

62 What about unit ball graph? Let x 1,..., x s be the consecutive centers in R 3 of a co-odd-cycle. Consider the trace on the 2-sphere of the following vector walk. Start at vector ab with a = x 1 and b = x 2. move continuously a from x 1 to x 3 following the segment x 1 x 3. move continuously b from x 2 to x 4 following the segment x 2 x 4. and so on, until back to x 1 x 2. As s is odd, half-way through we reach x 2 x 1. Hence the curve drawn on the 2-sphere is antipodal.

63 As two antipodal curves intersect, we have one of the following configurations: x y x y j x i y j x i y

64 Open questions Is Max Clique NP-hard on disk and unit ball graphs? A first step might be to show NP-hardness for Max Independent Set with iocp 1. Actually what about ocp 1? What is the complexity of Max Independent Set on the Moebius grid? on quadrangulations of the projective plane?

65 Open questions Is Max Clique NP-hard on disk and unit ball graphs? A first step might be to show NP-hardness for Max Independent Set with iocp 1. Actually what about ocp 1? What is the complexity of Max Independent Set on the Moebius grid? on quadrangulations of the projective plane? Thank you for your attention!