Maximum Clique on Disks - PDF Free Download

Maximum Clique on Disks 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 20 mars 2018, LIGM

Find a largest collection of disks that pairwise intersect

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Polynomial algorithm on unit disks by Clark, Colbourn, and Johnson, 1990. c i c j Guess two farthest disks in an optimum solution S.

Polynomial algorithm on unit disks by Clark, Colbourn, and Johnson, 1990. c i c j Hence, all the centers of S lie inside the bold digon.

Polynomial algorithm on unit disks by Clark, Colbourn, and Johnson, 1990. c i c j Two disks centered in the same-color region intersect.

Polynomial algorithm on unit disks by Clark, Colbourn, and Johnson, 1990. c i c j We solve Max Clique in a co-bipartite graph.

Polynomial algorithm on unit disks by Clark, Colbourn, and Johnson, 1990. c i c j We solve Max Independent Set in a bipartite graph.

Disk graphs Inherits the NP-hardness of planar graphs.

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

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)....

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

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.

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.

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

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.

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.

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.

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

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 =

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:

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

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

Complement of an even cycle 1, 2,..., 2s D 1 p 1 p p s p 1 p 2 s p p 3 p ps-2 s-1 4... D 1 p s-3 D 2 D 2s We start by positioning D 1, D 2, D 2s. We draw a convex chain between p 1 and p s.

Complement of an even cycle 1, 2,..., 2s D 1 p 1 p p s p 1 p 2 s p p 3 p ps-2 s-1 4... D 1 p s-3 D 2 D 2s We start by positioning D 1, D 2, D 2s. We draw a convex chain between p 1 and p s. D 2i : same radius and boundary crosses p i with tangent l(p i 1 p i+1 )

Stacking complements of even cycles D 2i+1 D 1 D 2i

Disks of different cycle complements intersect D 1 D 2i

Complement of odd cycle by unit disks (Atminas & Zamaraev)

Different representation with non-unit disks D 2i+1 D 1 D 2s+1 D 2i Same construction except D 1 intersects D 2s and D 2s+1 is "co-tangent" to D 1 and D 2s.

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

Sanity check: trying to stack complements of odd cycles D 2i+1 D 1 D 2s+1 D 2s +1 D 2i D 2s +1 cannot possibly intersect D 1

Going back to algorithms. Can we solve Max Independent Set more efficiently if there are no two vertex-disjoint odd cycles as an induced subgraph?

Going back to algorithms. 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

Quasi-polynomial time approximation-scheme (QPTAS) ocp(g): maximum size of an odd cycle packing. Theorem (Bock et al. 2014) PTAS for Max Independent Set for ocp = o(n/ log n).

Quasi-polynomial time approximation-scheme (QPTAS) ocp(g): maximum size of an odd cycle packing. Theorem (Bock et al. 2014) PTAS for Max Independent Set for ocp = o(n/ log n). Lemma Let H complement of a disk graph with n vertices. If ocp(h) > n/ log 2 n, then vertex of degree at least n/ log 4 n. Proof. The shortest odd cycle has size at most log 2 n.

Subexponential algorithm Theorem (Györi et al. 1997) A graph of odd girth c has an odd cycle cover of size Õ(n/c).

Subexponential algorithm Theorem (Györi et al. 1997) A graph of odd girth c has an odd cycle cover of size Õ(n/c). Let H be a complement of disk graph, its degree, c its odd girth. We can: branch in time 2Õ(n/ ). solve in time 2 O( c). solve in time 2Õ(n/c).

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

Filled triangles

Filled ellipses

The QPTAS and subexponential algorithm only work for intersection graphs of 2D-objects that are exactly disks. Approximation scheme with better running time? Have higher-dimensional disks the same obstruction?

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.

First step of the algorithm: sampling I is a fixed maximum solution. G I Select randomly a subset S of Õ(1/ε 3 ) vertices.

First step of the algorithm: sampling I is a fixed maximum solution. G I S With probability 1/2Õ(1/ε3), S I.

First step of the algorithm: sampling I is a fixed maximum solution. G N(S) I S Remove the neighborhood of S.

Classic result of Haussler and Welzl in VC dimension theory Theorem 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 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.

Second step of the algorithm: win-win on odd girth Now we removed N(S), all the vertices have degree < ε 3 I in I.

Second step of the algorithm: win-win on odd girth Now we removed N(S), all the vertices have degree < ε 3 I in I. We compute a shortest odd cycle C. If C 1/ε 2, we remove N[C] from the graph. The resulting graph is bipartite and N(C) < ε3 I = ε I. ε 2 If C > 1/ε 2, we need a fresh slide.

Finding a small odd cycle transversal Columns are the successive neighborhoods of C, layers. Rows indicate the closest vertex on C, strata.

Finding a small odd cycle transversal bipartite We delete the lightest among the first 1/ε layers.

Finding a small odd cycle transversal bipartite bipartite 1/ε consecutive strata are an odd cycle transversal.

Regarding the higher-dimensional question: 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 ))

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

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?

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? Thank you for your attention!