Computing a maximum clique in geometric superclasses of disk graphs
aa r X i v : . [ c s . C G ] J u l Computing a maximum clique in geometricsuperclasses of disk graphs
Nicolas Grelier ⋆ Department of Computer Science, ETH Zürich, Switzerland [email protected]
Abstract.
In the 90’s Clark, Colbourn and Johnson wrote a seminal pa-per, where they proved that maximum clique can be solved in polynomialtime in unit disk graphs. Since then, the complexity of maximum clique inintersection graphs of (unit) d -dimensional balls has been investigated.For ball graphs, the problem is NP-hard, as shown by Bonamy et al. (FOCS ’18). They also gave an efficient polynomial time approximationscheme (EPTAS) for disk graphs, however the complexity of maximumclique in this setting remains unknown. In this paper, we show the ex-istence of a polynomial time algorithm for solving maximum clique ina geometric superclass of unit disk graphs. Moreover, we give partialresults toward obtaining an EPTAS for intersection graphs of convexpseudo-disks. Keywords: pseudo-disks, line transversals, intersection graphs
In an intersection graph , every vertex can be represented as a set, such thattwo vertices are adjacent if and only if the corresponding sets intersect. In mostsettings, those sets are geometric objects, lying in a Euclidean space of dimen-sion d . Due to their interesting structural properties, the intersection graphs of d -dimensional balls, called d -ball graphs, have been extensively studied. For di-mensions , and , the d -ball graphs are called interval graphs , disk graphs and ball graphs , respectively. If all d -balls have the same radius, their intersectiongraphs are referred to as unit d -ball graphs . The study of these classes has manyapplications ranging from resource allocation to telecommunications [2, 9, 14].Many problems that are NP-hard for general graphs remain NP-hard for d -ball graphs, with fixed d ≥ . Even for unit disk graphs, most problems arestill NP-hard. A famous exception to this rule is the problem of computing amaximum clique, which can be done in polynomial time in unit disk graphsas proved by Clark, Colbourn and Johnson [7]. Their algorithm requires theposition of the unit disks to be given, but a robust version of their algorithm,which does not require this condition, was found by Raghavan and Spinrad [13]. ⋆ Research supported by the Swiss National Science Foundation within the collabora-tive DACH project Arrangements and Drawings as SNSF Project 200021E-171681. N. Grelier
This is a nontrivial matter as Kang and Müller have shown that the recognitionof unit d -ball graphs is NP-hard, and even ∃ R -hard, for any fixed d ≥ [12].Finding the complexity of computing a maximum clique in general diskgraphs (with arbitrary radii) is a longstanding open problem. However in 2017,Bonnet et al. , found a subexponential algorithm and a quasi polynomial time ap-proximation scheme (QPTAS) for computing a maximum clique on disk graphs [5].The following year, Bonamy et al. extended the result to unit ball graphs, andgave a randomised EPTAS for both settings [4]. The current state-of-the-artabout the complexity of computing a maximum clique in d -ball graphs is sum-marised in Table 1. unit d -ball graphs general d -ball graphs d = 1 linear [3] polynomial [10] d = 2 polynomial [7] Unknown but EPTAS [4, 5] d = 3 Unknown but EPTAS [4] NP-hard [4] d = 4 NP-hard [4] NP-hard [4]
Table 1.
Complexity of computing a maximum clique on d -ball graphs Bonamy et al. show that the existence of an EPTAS follows from the factthat: For any graph G that is a disk graph or a unit ball graph, the disjointunion of two odd cycles is a forbidden induced subgraph in the complementof G . Surprisingly, the proofs for disk graphs on one hand and unit ball graphson the other hand are not related. Bonamy et al. ask whether there is a naturalexplanation of this common property. They say that such an explanation couldbe to show the existence of a geometric superclass of disk graphs and unit ballgraphs, for which there exists an EPTAS for solving maximum clique.By looking at Table 1, a pattern seems to emerge: The complexity of com-puting a maximum clique on ( d − -ball graphs and unit d -ball graphs might berelated. We extend the question of Bonamy et al. and ask for a geometric inter-section graphs class that contains all interval graphs and all unit disk graphs, forwhich maximum clique can be solved in polynomial time. Recall that finding thecomplexity of maximum clique in disk graphs is still open. Therefore a secondmotivation for our question is that showing the existence of polynomial timealgorithms for geometric superclasses of unit disk graphs, may help to determinethe complexity of maximum clique in disk graphs.We introduce a class of geometric intersection graphs which contains all in-terval graphs and all unit disk graphs, for which we show that maximum cliquecan be solved efficiently. Furthermore, the definition of our class generalises toany dimension, i.e. for any fixed d ≥ we give a class of geometric intersectiongraphs that contains all ( d − -ball graphs and all unit d -ball graphs. We con-jecture that for d = 3 , there exists an EPTAS for computing a maximum cliquein the corresponding class. It is necessary that these superclasses be defined asclasses of geometric intersection graphs. Indeed, it must be if we want to un-derstand better the reason why efficient algorithms exist for both settings. For aximum clique in geometric graphs 3 instance, taking the union of interval graphs and unit disk graphs would notgive any insight, since it is a priori not defined by intersection graphs of somegeometric objects.In order to define the class, we first introduce the concept of d -pancakes . A -pancake is defined as the union of all unit disks whose centres lie on a linesegment s , with s itself lying on the x -axis. An example is depicted in Figure 1.This is definition is equivalent to the Minkowski sum of a unit disk centred atthe origin and a line segment on the x -axis, where the Minkowski sum of twosets A, B is equal to { a + b | a ∈ A, b ∈ B } . Similarly a -pancake is the union ofall unit balls whose centres lie on a disk D , with D lying on the xy -plane. Moregenerally, we have: x -axis s •• Fig. 1.
The union of the unit disks centred at points of s is a -pancake. Definition 1. A d -pancake is a d -dimensional geometric object. Let us denoteby { ξ , ξ , . . . , ξ d } the canonical basis of R d . A d -pancake is defined as theMinkowski sum of the unit d -ball centred at the origin and a ( d − -ball inthe hyperspace induced by { ξ , ξ , . . . , ξ d − } .We denote by Π d the class of intersection graphs of some finite collection of d -pancakes and unit d -balls. In this paper, we give a polynomial time algorithmfor solving maximum clique in Π : the intersection graphs class of unit disks and -pancakes. This is to put in contrast with the fact that computing a maximumclique in intersection graphs of unit disks and axis-parallel rectangles (insteadof -pancakes) is NP-hard and even APX-hard, as shown together with Bonnetand Miltzow [6], even though maximum clique can be solved in polynomial timein axis-parallel rectangle graphs [11].Relatedly, it would be interesting to generalise the existence of an EPTAS formaximum clique to superclasses of disk graphs. This was achieved with Bonnetand Miltzow for intersection graphs of homothets of a fixed bounded centrallysymmetric convex set [6]. In this paper, we aim at generalising further to inter-section graphs of convex pseudo-disks, for which we conjecture the existence ofan EPTAS, and give partial results towards proving it. The proof of these par-tial results relies on geometric permutations of line transversals. We do a caseanalysis on the existence of certain geometric permutations, and show that someconvex pseudo-disks must intersect. Holmsen and Wenger have written a survey N. Grelier on geometric transversals [16]. The results that are related to line transversalsare either of Hadwiger-type, concerned with the conditions of existence of linetransversals, or about the maximum number of geometric permutations of linetransversals. To the best of our knowledge, we do not know of any result thatuses geometric permutations of line transversals to show something else. Weconsider this tool, together with the polynomial time algorithm for computing amaximum clique in Π , to be our main contributions. Let G be a simple graph. We say that two vertices are adjacent if there isan edge between them, otherwise they are independent . For a vertex v , N ( v ) denotes its neighbourhood , i.e. the set of vertices adjacent to v . We denote by ω ( G ) , α ( G ) , and χ ( G ) the clique number, the independence number and thechromatic number of G , respectively.We denote by V ( G ) the vertex set of G . Let H be a subgraph of G . Wedenote by G \ H the subgraph induced by V ( G ) \ V ( H ) . We denote by G the complement of G , which is the graph with the same vertex set, but where edgesand non-edges are interchanged. A bipartite graph is graph whose vertex set canbe partitioned into two independent sets. A graph is cobipartite if its complementis a bipartite graph.We denote by iocp ( G ) the induced odd cycle packing number of G , i.e. themaximum number of vertex-disjoint induced odd cycles (for each cycle the onlyedges are the ones making the cycle), such that there is no edge between twovertices of different cycles. Throughout the paper we only consider Euclidean spaces with the Euclideandistance. Let p and p ′ be two points in R d . We denote by ( p, p ′ ) the line goingthrough them, and by [ p, p ′ ] the line segment with endpoints p and p ′ . We denoteby d ( p, p ′ ) the distance between them. For any fixed d , we denote by O the originin R d . When d = 2 , we denote by Ox and Oy the x and y -axis, respectively. For d = 3 , we denote by xOy the xy -plane. We usually denote a d -pancake by P d .As a reminder, a -pancake is the Minkowski sum of the unit disk centred at theorigin O and a line segment lying on the axis Ox . Definition 2.
Let { S i } ≤ i ≤ n be a family of subsets of R d . We denote the in-tersection graph of { S i } by G ( { S i } ) . It is the graph whose vertex set is { S i | ≤ i ≤ n } and where there is an edge between two vertices if and only if thecorresponding sets intersect. Definition 3. In R we denote by D ( c, ρ ) a closed disk centred at c with radius ρ . Let D = D ( c, ρ ) and D ′ = D ( c ′ , ρ ′ ) be two intersecting disks. We call lens aximum clique in geometric graphs 5 induced by D and D ′ the surface D ∩ D ′ . We call half-lenses the two closedregions obtained by dividing the lens along the line ( c, c ′ ) .For any x ≤ x , we denote by P ( x , x ) the -pancake that is the Minkowskisum of the unit disk centred at O and the line segment with endpoints x and x .Therefore we have P ( x , x ) = S x ≤ x ′ ≤ x D (( x ′ , , . Behind the definition ofthe d -pancakes is the idea that they should be the most similar possible to unit d -balls. In particular -pancakes should behave as much as possible like unitdisks. This is perfectly illustrated when the intersection of a -pancake and aunit disk is a lens, as the intersection of two unit disks would be. Definition 4.
Let { P j } ≤ j ≤ n be a set of -pancakes. For any unit disk D , wedenote by L ( D , { P j } ) , or simply by L ( D ) when there is no risk of confusion, theset of -pancakes in { P j } whose intersection with D is a lens.Let D denote D ( c, for some point c . Remark that if a -pancake P ( x , x ) for some x ≤ x is in L ( D ) , then the intersection between D and P ( x , x ) is equal to D ∩ D (( x , , or D ∩ D (( x , , . We make an abuse of notationand denote by d ( D , P ( x , x )) the smallest distance between c and a pointin the line segment [ x , x ] . Observe that if the intersection between D and P ( x , x ) is equal to D ∩ D (( x , , , then d ( D , P ( x , x )) = d ( c, ( x , ,and otherwise d ( D , P ( x , x )) = d ( c, ( x , . The following observation gives acharacterisation of when the intersection between a unit disk and a -pancake isa lens. Observation 1.
Let D (( c x , c y ) , be a unit disk intersecting with a -pancake P ( x , x ) . Then their intersection is a lens if and only if ( c x ≤ x or c x ≥ x ) and the interior of D (( c x , c y ) , does not contain any point in { ( x , ± , ( x , ± } . The observation follows immediately from the fact that the intersection is alens if and only if D (( c x , c y ) , does not contain a point in the open line segmentbetween the points ( x , − and ( x , − , nor in the open line segment betweenthe points ( x , and ( x , . We answer in Section 4 the -dimensional version of the question asked byBonamy et al. [4]: We present a polynomial time algorithm for computing a max-imum clique in a geometric superclass of interval graphs and unit disk graphs. Theorem 1.
There exists a polynomial time algorithm for computing a maxi-mum clique on Π , even without a representation. Kang and Müller have shown that for any fixed d ≥ , the recognition of unit d -ball graphs is NP-hard, and even ∃ R -hard [12]. We conjecture that it is alsohard to test whether a graph is in Π d for any fixed d ≥ , and prove it for d = 2 . Theorem 2.
Testing whether a graph is in Π is NP-hard, and even ∃ R -hard. N. Grelier
The proof of Theorem 2 figures in Section B in the Appendix. It immediatelyimplies that given a graph G in Π , finding a representation of G with -pancakesand unit disks is NP-hard. Therefore having a robust algorithm as defined in [13]is of interest. The algorithm of Theorem 1 takes any abstract graph as input,and outputs a maximum clique or a certificate that the graph is not in Π .For d = 3 , we conjecture the following: Conjecture 1.
There exists an integer K such that for any graph G in Π , wehave iocp ( G ) ≤ K .We show in Section C in the Appendix that this would be sufficient to obtainan EPTAS. Theorem 3.
If Conjecture 1 holds, there exists a randomised EPTAS for com-puting a maximum clique in Π , even without a representation. By construction the class Π d contains all ( d − -ball graphs and all unit d -ball graphs. Indeed a ( d − -ball graph can be realised by replacing in arepresentation each ( d − -ball by a d -pancake. In addition to this property,we want fast algorithms for maximum clique on Π d . The definition of Π d mayseem unnecessarily complicated. The most surprising part of the definition isprobably the fact that we use d -pancakes instead of simply using ( d − -ballsrestricted to be in the same hyperspace of R d . However, we show in Section Ain the Appendix that our arguments for proving fast algorithms would not holdwith this definition.We give partial results toward showing the existence of an EPTAS for maxi-mum clique on intersection graphs of convex pseudo-disks . We say that a graphis a convex pseudo-disk graph if it is the intersection graph of convex sets in theplane such that the boundaries of every pair intersect at most twice. We denoteby G the intersection class of convex pseudo-disk graphs. A structural propertyused to show the existence of an EPTAS for disk graphs is that for any diskgraph G , iocp ( G ) ≤ . The proof of Bonnet et al. relies heavily on the fact thatdisks have centres [5]. However, convex pseudo-disks do not, therefore adaptingthe proof in this new setting does not seem easy. While we were not able toextend this structural result to the class G , we show a weaker property: Thecomplement of a triangle and an odd cycle is a forbidden induced subgraph in G .We write “complement of a triangle" to make the connection with iocp clear,but remark that actually the complement of a triangle is an independent set ofthree vertices. Below we state this property more explicitly. Theorem 4.
Let G be in G . If there exists an independent set of size , denotedby H , in G , and if for any u ∈ H and v ∈ G \ H , { u, v } is an edge of G , then G \ H is cobipartite. Note that a cobipartite graph is not the complement of an odd cycle. Giventhe three pairwise non-intersecting convex pseudo-disks in H , we give a geometriccharacterisation of the two independent sets in the complement of G \ H . Weconjecture that Theorem 4 is true even when H is the complement of any odd-cycle, which implies: aximum clique in geometric graphs 7 Conjecture 2.
For any convex pseudo-disk graph G , we have iocp ( G ) ≤ .If Conjecture 2 holds, it is straightforward to obtain an EPTAS for maximumclique in convex pseudo-disks graphs, by using the techniques of Bonamy etal. [4]. Π in polynomialtime In this section we prove Theorem 1. The idea of the algorithm is similar to theone of Clark, Colbourn and Johnson [7]. We prove that if u and v are the mostdistant vertices in a maximum clique, then N ( u ) ∩ N ( v ) is cobipartite.In their proof, Clark, Colbourn and Johnson use the following fact: if c and c ′ are two points at distance ρ , then the diameter of the half-lenses induced by D ( c, ρ ) and D ( c ′ , ρ ) is equal to ρ . We prove here a similar result. Lemma 1.
Let c and c ′ be two points at distance ρ , and let be ρ ′ ≥ ρ . Then thediameter of the half-lenses induced by D ( c, ρ ) and D ( c ′ , ρ ′ ) is at most ρ ′ .Proof. First note that if ρ ′ > ρ then the half-lenses are half-disks of D ( c, ρ ) .The diameter of these half-disks is equal to ρ , which is smaller than ρ ′ . Let usnow assume that we have ρ ′ ≤ ρ . The boundary of the lens induced by D ( c, ρ ) and D ( c ′ , ρ ′ ) consists of two arcs. The line ( c, c ′ ) intersects exactly once witheach arc. One of these two intersections is c ′ , we denote by c ′′ the other. Letus consider the disk D ( c ′′ , ρ ′ ) . Note that it contains the disk D ( c, ρ ) . Thereforethe lens induced by D ( c, ρ ) and D ( c ′ , ρ ′ ) is contained in the lens induced by D ( c ′′ , ρ ′ ) and D ( c ′ , ρ ′ ) , whose half-lenses have diameter ρ ′ . The claim followsfrom the fact that the half-lenses of the first lens are contained in the ones ofthe second lens.Before stating the next lemma, we introduce the following definition: Definition 5.
Let { S i } ≤ i ≤ n and { S ′ j } ≤ j ≤ n ′ be two families of sets in R . Wesay that { S i } and { S ′ j } fully intersect if for all i ≤ n and j ≤ n the intersectionbetween S i and S ′ j is not empty. Lemma 2.
Let D := D ( c, be a unit disk and let P := P ( x , x ) be in L ( D ) .Let {D i } be a set of unit disks that fully intersect with {D , P } , such that forany D i we have d ( D , D i ) ≤ d ( D , P ) . Moreover if P is in L ( D i ) we require d ( D i , P ) ≤ d ( D , P ) . Also let { P j } be a set of -pancakes that fully intersectwith {D , P } , such that for any P j in { P j }∩ L ( D ) , we have d ( D , P j ) ≤ d ( D , P ) .Then G ( {D i } ∪ { P j } ) is cobipartite.Proof. The proof is illustrated in Figure 2. Without loss of generality, let usassume that the intersection between D and P is equal to D ∩ D (( x , , .Remember that by definition we have x ≤ x . Let P ( x ′ , x ′ ) be a -pancake in { P j } . As it is intersecting with P , we have x ′ ≥ x − . Assume by contradiction N. Grelier that we have x ′ > x . Then with Observation 1, we have that P ( x ′ , x ′ ) is in L ( D ) and d ( D , P ( x ′ , x ′ )) > d ( D , P ) , which is impossible. Therefore we have x ′ ≤ x , and so P ( x ′ , x ′ ) must contain D (( x ′ , , for some x ′ satisfying x − ≤ x ′ ≤ x . As the line segment [( x − , , ( x , has length , the -pancakes in { P j } pairwise intersect.We denote by ρ the distance d ( D , P ) . Let D ( c i , be a unit disk in {D i } .By assumption, c i is in D ( c, ρ ) ∩ D (( x , , . We then denote by R the lensthat is induced by D ( c, ρ ) and D (( x , , . We cut the lens into two parts withthe line ( c, ( x , , and denote by R the half-lens that is not below this line,and by R the half-lens that is not above it. With Lemma 1, we obtain that thediameter of R and R is at most . Let us assume without loss of generalitythat c is not below Ox . We denote by X the set of unit disks in {D i } whosecentre is in R . We denote by X the union of { P j } and of the set of unitdisks in {D i } whose centre is in R . Since the diameter of R is , any pair ofunit disks in X intersect, therefore G ( X ) is a complete graph. To show that G ( X ) is a complete graph too, it remains to show that any unit disk D ( c i , in X and any -pancake P ( x ′ , x ′ ) in { P j } intersect. We denote by P thefollowing convex shape: ∪ x ′ ≤ x ≤ x ′ D (( x, , . Note that the fact that D ( c i , and P ( x ′ , x ′ ) intersect is equivalent to having c i in P . Let us consider thehorizontal line going through c , and let us denote by c ′ the left intersection withthe circle centred at ( x , with radius . We also denote by r the extremity of R that is in R .Let us assume by contradiction that c i is above the line segment [ c, c ′ ] . Asby assumption c i is in R , it implies that the x -coordinate of c i is smaller thanthe one of c . Therefore P is in L ( D i ) and d ( D i , P ) > d ( D , P ) , which isimpossible by assumption. Let us denote by R , − the subset of R that is notabove the line segment [ c, c ′ ] . To prove that D ( c i , and P ( x ′ , x ′ ) intersect,it suffices to show that P contains R , − . As shown above, P ( x ′ , x ′ ) contains D (( x ′ , , for some x ′ satisfying x − ≤ x ′ ≤ x . This implies that P contains D (( x − , , ∩ D (( x , , , and in particular contains x . Moreover as c isnot below Ox , r is also in D (( x − , , ∩ D (( x , , . As P intersects D , P contains c . Let us assume by contradiction that P does not contain c ′ . Then x ′ must be smaller than the x -coordinate of c ′ , because otherwise the distance d (( x ′ , , c ′ ) would be at most d (( x , , c ′ ) , which is equal to . But then if P does not contain c ′ , then it does not contain c either, which is a contradiction.We have proved that P contains the points x , c , c ′ and r . By convexity, andusing the fact that two circles intersect at most twice, we obtain that R , − iscontained in P . This shows that any two elements in X intersect, which impliesthat G ( X ) is a complete graph. Finally, as X ∪ X = {D i } ∪ { P j } , we obtainthat G ( {D i }∪{ P j } ) can be partitioned into two cliques, i.e. it is cobipartite. Lemma 3.
Let D := D (( c x , c y ) , and D ′ := D (( c ′ x , c ′ y ) , be two unit diskssuch that c x ≤ c ′ x . Let P := P ( x , x ) be a -pancake intersecting with D and D ′ , such that x ≥ c x and P is not in L ( D ) . Let P := P ( x ′ , x ′ ) be a aximum clique in geometric graphs 9 ( x , ( x − , cc ′ r R R D P •• ••• Fig. 2.
Illustration of the proof of Lemma 2 -pancake intersecting with D and D ′ , but not intersecting with P , then P isin L ( D ) ∩ L ( D ′ ) .Proof. The proof is illustrated in Figure 3. First let us prove that P cannot be onthe right side of P , i.e. we have x ′ ≤ x . Otherwise we would have x ′ ≥ x , andthen as P and P are not intersecting, we would have x ′ > x + 2 . Hence, sincewe assume x ≥ c x , we obtain d ( c, ( x ′ , > , which is impossible. Thereforewe have x ′ ≤ x , and even x ′ < x − since P and P are not intersecting.Without loss of generality, let us assume ≤ c y . Let us consider the horizontalline ℓ with height . By assumption it intersects with the circle centred at ( c x , c y ) with unit radius. There are at most two intersections, and we denote by x ℓ the x -coordinate of the one the right. As P is not in L ( D ) , we have x ≤ x ℓ . Then,since x ′ < x − and the fact that D has diameter , we know that the points ( x ′ , and ( x ′ , − are not in D , which implies that P is in L ( D ) . Likewise aswe have c x ≤ c ′ x , the points ( x ′ , and ( x ′ , − are not in D ′ , and so P is in L ( D ) ∩ L ( D ′ ) . Lemma 4.
Let D := D ( c, and D ′ := D ( c ′ , be two intersecting unit disks.Let {D i } be a set of unit disks that fully intersect with {D , D ′ } , such that foreach unit disk D i we have d ( D , D i ) ≤ d ( D , D ′ ) and d ( D ′ , D i ) ≤ d ( D , D ′ ) . Also let { P j } be a set of -pancakes that fully intersect with {D , D ′ } , such that for any P j in { P j } ∩ L ( D ) , we have d ( D , P j ) ≤ d ( D , D ′ ) , and for any P j in { P j } ∩ L ( D ′ ) ,we have d ( D ′ , P j ) ≤ d ( D , D ′ ) . Then G ( {D i } ∪ { P j } ) is cobipartite.Proof. We denote by ρ the distance between c and c ′ . We denote by R the lensinduced by D ( c, ρ ) and D ( c ′ , ρ ) . We cut R with the line segment [ c, c ′ ] , whichpartitions R into two half-lenses that we denote by R and R . By assumption,the centre of any unit disk in {D i } must be in R . Since R and R have diameter ρ which is at most , any two unit disks having their centre in R must intersect,and the same holds with R . Therefore G ( {D i } ) is cobipartite, which is the claimif { P j } is empty. ( x ℓ , x ℓ − , D D ′ Oxℓ •• •
Fig. 3.
Illustration of the proof of Lemma 3
We now assume that { P j } is not empty. In order to show the claim, we do acase analysis according to whether the intersection between D ( c, ∩D ( c ′ , and Ox is empty or not. Let us assume that the latter holds, as shown in Figure 4.Let P be in a -pancake in { P j } . By convexity, P contains a unit disk centredat ( x ′ , , where ( x ′ , is in D ( c, ∩ D ( c ′ , . We denote by R + the lens thatinduced by D ( c, and D ( c ′ , and cut it with the line ( c, c ′ ) . We denote by R +1 (respectively R +2 ) the half-lens that contains R (respectively R ). Let usassume that ( x ′ , is in R +1 . By assumption D (( x ′ , , contains c and c ′ . Let usconsider the third extremity of R , along with c and c ′ , that we denote by r . Bymaking a circle centred at r grow, we observe that the farthest point from r in R +1 can only be at one of the three extremities of R +1 . However by Lemma 1 thesedistances are at most , which implies that the distance between ( x ′ , and r isat most . Using the fact that two circles intersect at most twice, we obtain that R is contained in D (( x ′ , , . Therefore P intersect with all unit disks whosecentre is in R , and with all -pancakes in L ( D ) ∩ L ( D ′ ) that contain a diskwhose centre is in R . Let P ( x , x ) and P ( x ′ , x ′ ) be two -pancakes in { P j } such that they contain each a unit disk whose centre is in R +1 , but such thatthey do not contain a unit disk whose centre is in R . In particular, P ( x , x ) and P ( x ′ , x ′ ) are not in L ( D ) ∩ L ( D ′ ) . We claim that they intersect. Supposeby contradiction that they do not. Without loss of generality, let us assume that P ( x , x ) is to the right of P ( x ′ , x ′ ) , and that c x ≤ c ′ x , where c x and c ′ x denotethe x -coordinate of c and c ′ respectively. Since P ( x , x ) does not contain a diskin R , and since it is on the right side of P ( x ′ , x ′ ) , it implies that it does notcontain a disk with centre in D ( c, ρ ) . Therefore P ( x , x ) cannot be in L ( D ) .Moreover the fact that it does not contain a disk with centre in D ( c, ρ ) implies x ≥ c x . We finally apply Lemma 3 to obtain a contradiction. We denote X the set of unit disks whose centre is in R and -pancakes that contain a diskwhose centre is in R +1 . We know that two unit disks in X intersect. Moreoverwe have shown that a -pancake and a unit disk in X intersect. For a pair of aximum clique in geometric graphs 11 two pancakes, if one of them contains a disk whose centre is in R it is done forthe same reasons. If none of them does, then we have shown above that theyintersect. This shows that G ( X ) is a complete graph. By defining X as the setof the remaining disks and -pancakes, using the symmetry of the problem weobtain that G ( X ) is also a complete graph.Now let us assume that the intersection between D ( c, ∩ D ( c ′ , and Ox isempty, as shown in Figure 5. As { P j } is not empty, the set Ox \ ( D ( c, ∪D ( c ′ , consists of three connected component, one of them being bounded. We denote by s the closed line segment consisting of the bounded connected component and itsboundaries. Any -pancake P in { P j } contains a point in D ( c, ∩ Ox , otherwise P would not intersect with D . Likewise P contains a point in D ′ ∩ Ox , andtherefore contains s . This implies that all -pancakes in { P j } pairwise intersect.Let us assume without loss of generality that R is closer to Ox than R . Letus show that any -pancake P in { P j } and any unit disk whose centre is in R intersect. This is equivalent to show that for any point p in R , there is a pointin P ∩ Ox that is at Euclidean distance at most from p . We denote by P the Minkowski sum of the disk with radius centred at O and the line segment s , i.e. P = ∪ x ′ ∈ s D ( x ′ , . Note that P is convex. We claim that P contains R , which implies the desired property. Since s contains a point p in D ( c, , weknow that P contains c . Likewise, as s contains a point p in D ( c ′ , , then P contains c ′ , and therefore by convexity the whole line segment [ c, c ′ ] . Therefore P contains the quadrilateral cc ′ p p . If this quadrilateral contains R we aredone. Otherwise, it may not contain a circular segment of the disk D ( c ′ , ρ ) ora circular segment of the disk D ( c, ρ ) . Let us assume that we have the worstcase, meaning that both circular segments are not in cc ′ p p . Let us considerthe circle C centred at p with radius , and the circle C ′ centred at c ′ withradius ρ . The two circles intersect at c . Let us consider the point p ′ that is atthe intersection between C ′ and the line segment [ c, p ] . By definition, p ′ is insidethe disk induced by C . As two circles intersect at most twice, we obtain thatthe arc cp ′ centred at c ′ with radius ρ is contained in the disk induced by C ,and therefore also in P . By convexity, we know that the circular segment ofthe disk D ( c ′ , ρ ) with chord [ c, p ′ ] is in P . We can apply the same argumentsfor the other side to show that R is in P . Hence by defining X as the set ofdisks whose centre centre is in R , union the set of -pancakes, and X as theset of disks whose centre is in R , we have that G ( X ) and G ( X ) are completegraphs.Note that Lemma 2 and Lemma 4 give a polynomial time algorithm for max-imum clique on Π when a representation is given. First compute a maximumclique that contains only -pancakes, which can be done in polynomial time sincethe intersection graph of a set of -pancakes is an interval graph [10]. Then foreach unit disk D , compute a maximum clique which contains exactly one unitdisk, D , and an arbitrary number of -pancakes. Because finding out whether aunit disk and a -pancake intersect takes constant time, computing such a max-imum clique can be done in polynomial time. Note that if a maximum cliquecontains at least two unit disks, then in quadratic time we can find in this max-imum clique either a pair of unit disks or a unit disk and a -pancake whoseintersection is a lens, such that the conditions of Lemma 2 or of Lemma 4 are Ox c c ′ r R R R +1 R +2 • •• Fig. 4.
First case: D ( c, ∩ D ( c ′ , ∩ Ox = ∅ c c ′ D ( c, D ( c ′ , Ox C p p ′ p R s • ••• • Fig. 5.
Second case: D ( c, ∩ D ( c ′ , ∩ Ox = ∅ satisfied. By applying the corresponding lemma, we know that we are comput-ing a maximum clique in a cobipartite graph, which is the same as computinga maximum independent set in a bipartite graph. As this can be done in poly-nomial time [8], we can compute a maximum clique on Π in polynomial timewhen the representation is given. To obtain an algorithm that does not require a representation, we use the no-tion of cobipartite neighborhood edge elimination ordering (CNEEO) as intro-duced by Raghavan and Spinrad [13]. Let G be a graph with m edges. Let Λ = e , e , . . . , e m be an ordering of the edges. Let G Λ ( k ) be the subgraph of G with edge set { e k , e k +1 , . . . , e m } . For each e k = ( u, v ) , N Λ,k is defined as the setof vertices adjacent to u and v in G Λ ( k ) . Definition 6 (Raghavan and Spinrad [13]).
An edge ordering Λ = { e , e , . . . , e m } is a CNEEO if for each e k , N Λ,k induces a cobipartite graph in G . Lemma 5 (Raghavan and Spinrad [13]).
Given a graph G and a CNEEO Λ for G , a maximum clique on G can be found in polynomial time. aximum clique in geometric graphs 13 They propose a greedy algorithm for finding a CNEEO: When having chosenthe first i − edges e , . . . , e i − , try every remaining edges one by one untilfinding one that satisfies the required property. If no such edge exists, returnthat the graph does not admit a CNEEO, which follows from Lemma 6. Lemma 6 (Raghavan and Spinrad [13]). If G admits a CNEEO, then thegreedy algorithm finds a CNEEO for G . To show that it is possible to compute a maximum clique on a graph G in Π , we show that such a graph admits a CNEEO. As noted by Raghavan andSpinrad, the algorithm computes a maximum clique for any graph that admitsa CNEEO, and otherwise states that the given graph does not admit a CNEEO.In particular, the algorithm does not say whether the graph is indeed in Π , andcannot be used for recognition. Theorem 5.
Let G be a graph in Π , then G admits a CNEEO. Theorem 5, Lemma 5 and Lemma 6 immediately imply Theorem 1. To proveTheorem 5, we use two more lemmas.
Lemma 7.
Let D = D (( c x , c y ) , be a unit disk. Let { P j } be a set of -pancakesthat all intersect with D . Then G ( { P j } ) is cobipartite.Proof. Let P ( x , x ) be in { P j } . By triangular inequality we have x ≤ c x +2 or x ≥ c x − . It implies that P ( x , x ) contains the line segment [( x ′ − , , ( x ′ +1 , for some x ′ satisfying c x − ≤ x ′ ≤ c x + 2 . We define X as the set of -pancakes in { P j } that contain the line segment [( x ′ − , , ( x ′ + 1 , for some x satisfying c x − ≤ x ′ ≤ c x , and X as { P j } \ X . We obtain that G ( X ) and G ( X ) are complete graphs. Lemma 8.
Let P = P ( x , x ) and P ′ = P ( x ′ , x ′ ) be two intersecting -pancakes. Let { P j } be a set of -pancakes that fully intersect with { P , P ′ } ,such that for any P j in { P j } , P j is not contained in P nor in P ′ . Then G ( { P j } ) is cobipartite.Proof. Let P ( x ′′ , x ′′ ) be in { P j } . Let us first assume that one of P , P ′ iscontained in the other. Without loss of generality, let us assume that P is con-tained in P ′ , which is equivalent to having x ′ ≤ x ≤ x ≤ x ′ . By assumption,as P ( i, j ) is not contained in P , we have x ′′ < x or x < x ′′ . As P ( x ′′ , x ′′ ) pairwise intersect with P and P ′ , it implies that P ( x ′′ , x ′′ ) contains ( x , or ( x , . We define X as the set of -pancakes in { P j } that contains ( x , , and X as { P j } \ X . We obtain that G ( X ) and G ( X ) are complete graphs.If none of P , P ′ is contained in the other, we can assume without loss ofgenerality that x ≤ x ′ ≤ x ≤ x ′ . Therefore we have x ′′ < x ′ or x < x ′′ , whichimplies that P ( x ′′ , x ′′ ) contains ( x ′ , or ( x , . We conclude as above. Proof of Theorem 5.
We divide out set of edges into three: E , E and E . E contains all the edges between a pair of unit disks, or between a unit disk D anda -pancake in L ( D ) . E contains the edges between a unit disk and a -pancakethat are not in E . E contains the edges between a pair of -pancakes. For anedge e = { u, v } in E , we called length of e the distance between u and v , bethey unit disks or a unit disk D and a -pancake in L ( D ) . We order the edges in E by non increasing length, which gives an ordering Λ . We take any ordering Λ of the edges in E . For E , we take any ordering Λ such that for any edge e = { u, v } , no edge after e in Λ contains a -pancake contained in u or v . Thiscan be obtained by considering the smallest -pancakes first. We finally define anordering Λ = Λ Λ Λ on E . Let us consider an edge e k . If e k is in E , Lemma 2and Lemma 4 show that N Λ,k induces a cobipartite graph. If e k is in E , we useLemma 7, and if e k is in E , we conclude with Lemma 8. This shows that Λ isa CNEEO. In this section we are interested in computing a maximum clique in intersectiongraphs of convex pseudo-disks. Our proof relies on line transversal and theirgeometric permutations on the three convex pseudo-disks that form a trianglein the complement, denoted by D , D and D . As there are only three sets, thegeometric permutation of a line transversal is given simply by stating which setis the second one intersected. Definition 7. A line transversal ℓ is a line that goes through the three convexpseudo-disks D , D and D . We call (convex pseudo-)disk in the middle of aline transversal the convex pseudo-disk it intersects in second position.For sake of readability, we from now on omit to mention that a disk in themiddle is a convex pseudo-disk, and simply refer to it as disk in the middle.We are going to conduct a case analysis depending on the number of convexpseudo-disks being the disk in the middle for some line transversal. When thereexists no line transversal, we can prove a stronger statement. Lemma 9.
If there is no line transversal through a family of convex sets F ,then for any pair of convex sets { C , C } that fully intersects with F , C and C intersect.Proof. Let us prove the contrapositive. Assume that C and C do not intersect,therefore there exists a separating line. As all sets in F intersect C and C , theyalso intersect the separating line, which is thus a line transversal of F .Using the notation of Theorem 4, Lemma 9 immediately implies that if thereis no line transversal through the sets representing H , then G \ H is a clique,which is an even stronger statement than required.The following lemma is the key-tool used in our proofs. It is illustrated inFigure 6. Let D , D and D be three convex pseudo-disks that do not pairwiseintersect. aximum clique in geometric graphs 15 p ′ p ′ p ′ p p p • •• •• • Fig. 6.
Illustration of Lemma 10. The triangles p p p and p ′ p ′ p ′ intersect. Lemma 10.
Assume that D nor D is the disk in the middle of a line transver-sal. Let p i and p ′ i be in D i , ≤ i ≤ . For sake of simplicity, assume that theline ( p , p ) is horizontal. If at least one of p ′ , p ′ is above this line, and D isbelow it, then the triangles p p p and p ′ p ′ p ′ intersect.Proof. Suppose by contradiction that the two triangles do not intersect. Thusthere is a separating line ℓ . The separating line intersects with [ p i , p ′ i ] , ≤ i ≤ ,and by convexity it is a line transversal of {D , D , D } . By assumption, itsintersection with D is below the line ( p , p ) . However, as one of p ′ , p ′ is above ( p , p ) , the part of ℓ between D and D is also above ( p , p ) . This implies thateither D or D is the disk in the middle of ℓ , which is a contradiction. Remarkthe lemma can be generalised to any convex set {D , D , D } , since we did notuse the fact that they are pseudo-disks.Let {D ′ j } be a set of convex pseudo-disks that fully intersect with {D , D , D } . Lemma 11.
If there exists one convex pseudo-disk D i ∈ {D , D , D } such thatthe disk in the middle of all line transversals of {D , D , D } is D i , then G ( {D ′ j } ) is cobipartite.Proof. Without loss of generality, let us assume that the disk in the middle ofall line transversals is D . Let ℓ be a line transversal, that we will assume tobe horizontal. Let D ′ be a convex pseudo-disk intersecting pairwise with D , D and D . We denote by p ′ a point in D ′ ∩ D and by p ′ a point in D ′ ∩ D . If theline segment [ p ′ , p ′ ] intersect D , we have the following: Since D ′ and D arepseudo-disks, then D ′ must either contain all the part of D that is above or theone that is below the line ( p ′ , p ′ ) . We will partition the convex pseudo-disks in {D ′ j } in four sets depending on the line segment [ p ′ , p ′ ] .1. [ p ′ , p ′ ] is above D ,2. [ p ′ , p ′ ] intersects D and D ′ contains all the part of D above it,3. [ p ′ , p ′ ] is below D ,4. [ p ′ , p ′ ] intersects D and D ′ contains all the part of D below it. We are going to show that the set X ⊆ {D ′ j } of convex pseudo-disks in case or all pairwise intersect. By symmetry, the same holds for the set X ⊆ {D ′ j } of convex pseudo-disks in cases and , and thus the claim will follow.Let us suppose that we have two convex pseudo-disks D ′ and D ′′ in case . Let us suppose by contradiction that they do not intersect. Without loss ofgenerality, let us assume that p ′ is above [ p ′′ , p ′′ ] . We can then apply Lemma 10to show that D ′ and D ′′ intersect, which is a contradiction.Let us assume that D ′ and D ′′ are in case . If the line segments [ p ′ , p ′ ] and [ p ′′ , p ′′ ] intersect then it is done by convexity. Therefore we can assume withoutloss of generality that [ p ′ , p ′ ] ∩ D is above [ p ′′ , p ′′ ] ∩ D . Hence both D ′ and D ′′ contain [ p ′ , p ′ ] ∩ D , which shows that they intersect.Finally, let us assume without loss of generality that D ′ is in case and D ′′ isin case . Suppose by contradiction that D ′ and D ′′ do not intersect. Thereforeby convexity the line segments [ p ′ , p ′ ] and [ p ′′ , p ′′ ] do not intersect. We deducethat [ p ′ , p ′ ] is above [ p ′′ , p ′′ ] ∩ D . Let p ′ be a point in D ′ ∩ D . By assumption p ′ cannot be above [ p ′′ , p ′′ ] , otherwise it would also be in D ′′ . As there is no linetransversal having D or D as disk in the middle, we apply Lemma 10 to showthat D ′ and D ′′ intersect, which concludes the proof. Definition 8.
Let D and ˜ D be two non intersecting disks and let p, q be in theinterior D , ˜ D respectively. We call external tangents of D and ˜ D the two onesthat do not cross the line segment [ p, q ] . D D ′′ D D D ′ ττ ′ S Fig. 7. D is contained, D ′ is -intersecting and D ′′ is -intersecting. Definition 9.
Let D i be a disk in {D , D , D } , such that it is the disk in themiddle of a line transversal. For sake of simplicity, let us assume that this disk is D . We denote by τ and τ ′ the two external tangents of D and D . We say that aximum clique in geometric graphs 17 D is contained if it is included in the surface S delimited by D , τ , D and τ .If the intersection between D and exactly one of the external tangents is notempty, we say that D is -intersecting . If the intersection with both externaltangents is not empty, we say that D is -intersecting . The different cases areillustrated in Figure 7. Lemma 12.
Let D i be a disk in {D , D , D } , such that it is the disk in themiddle of a line transversal. If D i is contained, then it is possible to split {D ′ j } into two sets X and X , such that G ( X ) and G ( X ) are complete graphs.Proof. Without loss of generality, let us that the considered disk is D . By as-sumption, there is a line transversal having D as disk in the middle. Let ℓ besuch a line transversal, and let us assume that is horizontal. Let D ′ be a diskthat pairwise intersect with D , D and D . Let p ′ be a point in D ′ ∩ D . Wechoose likewise p ′ and p ′ . We will distinguish different cases. We define onlythree, the others can be obtained by symmetry by replacing "above" by "below"and vice versa.1. [ p ′ , p ′ ] is above D , and ([ p ′ , p ′ ] ∩ D = ∅ ) ∩ ([ p ′ , p ′ ] ∩ D = ∅ ) ,2. [ p ′ , p ′ ] intersects with D and D ′ contains all the part of D above it,3. [ p ′ , p ′ ] is below D , and ([ p ′ , p ′ ] ∩ D = ∅ ) ∪ ([ p ′ , p ′ ] ∩ D = ∅ ) .We are going to prove that the disks in the set X of all disks in case , or , pairwise intersect. The same also holds for X where we replace "above" by"below" and vice versa in the definitions. Let D ′ and D ′′ be two disks in X . Ifboth correspond to case or , we can apply the same reasoning as in cases and of Lemma 11 to show that they intersect. For a disk in case we cannotuse the fact that there is no line transversal, but as we have ([ p ′ , p ′ ] ∩ D = ∅ ) ∪ ([ p ′ , p ′ ] ∩ D = ∅ ) , we can still apply Lemma 10.Let D ′ be a disk in case . If we can find If D ′ is in case and D ′′ in case , Lemma 13.
If a disk D i is -intersecting, then it is the disk in the middle ofall line transversals.Proof. Without loss of generality let us assume that this disk is D . By definition,it is the disk in the middle of a line transversal. We denote by τ and τ ′ theexternal tangents. let p be a point in D ∩ τ and p ′ be in D ∩ τ ′ . The linesegment [ p, p ′ ] is included in D , and separates D from D . Let ℓ be a linetransversal. Let p be in ℓ ∩ D and p be in ℓ ∩ D . The line segment [ p , p ] must cross [ p, p ′ ] , which shows that the disk in the middle of ℓ is D . Definition 10.
Let D i be a -intersecting disk. We denote by τ i the externaltangents of the two other disks that D i intersects. We denote by A i the partof D i that is on the same side of τ i as the two other disks. It is illustrated inFigure 8. τ D D D A Fig. 8.
Illustration of Definition 10, with D being the -intersecting disk τ D D D p ′ p ′ p ′ p ′′ p ′′ • •• •• Fig. 9.
Illustration of the proof of Lemma 14.
Lemma 14.
If there exists exactly one disk, say D , such that there is notransversal having D as disk in the middle, then it is possible to split {D ′ j } into two sets X and X , such that G ( X ) and G ( X ) are complete graphs.Proof. There is a line transversal having D as disk in the middle. If D is con-tained then we apply Lemma 12 to conclude. Otherwise we know with Lemma 13that D is -intersecting. We use the notation as in Definition 10. Without lossof generality, let us assume that τ is horizontal, and that both D and D arebelow it. As D is not contained, D \ A is not empty. We denote by X theset of all disks in {D ′ j } that contain D \ A . Let X be the set of the remainingdisks. By definition, two disks in X intersect. Therefore G ( X ) is a completegraph.Let D ′ and D ′′ be two disks in X . The following proof is illustrated inFigure 9. Let p ′ be a point in D ′ ∩ D and p ′ in D ′ ∩ D , and likewise for p ′′ and p ′′ . If the two line segments [ p ′ , p ′ ] and [ p ′′ , p ′′ ] intersect then it is done.Otherwise let us assume without loss of generality that [ p ′ , p ′ ] is above [ p ′′ , p ′′ ] .Let p ′ be a point in D ′ ∩ D . As there is no line transversal with D as disk inthe middle, we have that [ p ′ , p ′ ] intersects with A , is above p ′′ , and that it doesnot cross τ . Therefore, since D ′ does not contain D \ A , it must contain all thepart of D that is below the line ( p ′ , p ′ ) . Hence D ′ contains p ′′ , which impliesthat D ′ and D ′′ intersect. We have shown that G ( X ) is a complete graph.The following definition is illustrated in Figures 10 and 11. Definition 11.
Let D i in {D , D , D } be a disk that is -intersecting, say D i = D . Let D ′ be a disk intersecting pairwise with D , D and D . We say that D ′ aximum clique in geometric graphs 19 D D D D ′ χ χ •• Fig. 10.
Illustration of Definition 11, D ′ is centred D D D D ′′ Fig. 11.
Illustration of Definition 11, D ′′ is not centred0 N. Grelier is outside-containing D if D \ A is a proper subset of D ′ . We denote by χ and χ the points where the boundaries of D ′ and D intersect. Note that theyare both in A . We denote by H the closed halfplane with bounding line ( χ , χ ) that contains D \ A . Let H ′ be the closed halfplane with bounding line τ thatcontains A . Note that ( H \ H ′ ) \ A is the union of two connected sets. Remarkthat we have D ′ ∩ D ⊂ H \ H ′ and D ′ ∩ D ⊂ H \ H ′ . If D ′ ∩ D and D ′ ∩ D are not in the same connected set, we say that D ′ is centred . Lemma 15. If D ′ and D ′′ are centred, then they intersect.Proof. Let D ′ and D ′′ be two disks in that are centred. If they both contain thesame surface D i \ A i , then they intersect. Otherwise, let us assume without lossof generality that D ′ contains D \ A and D ′′ contains D \ A . There are twointersections between τ and the boundary of D . Let p ′′ be the intersection thatis the closest to D . Let χ and χ be the points where the boundaries of D ′ and D intersect, such that χ is the closest to D . We denote by H the closedhalfplane with bounding line ( χ , χ ) containing D \ A Let p ′′ be in D ′′ ∩ D .If p ′′ is in D ∩ H then it is done. Let us assume that it is not. Let p ′ be in D ′ ∩ D . Then the line segments [ χ , p ′ ] and [ p ′′ , p ′′ ] intersect or p ′ is in D \ A .In both case D ′ and D ′′ intersect. Lemma 16. If D ′ is not centred and D ′′ is any disk not centred, then theyintersect. Remark that in Lemma 16, D ′′ is not required to be outside-containing somedisk. Proof.
Without loss of generality, let us assume that D ′ is outside-containing D .We denote by χ and χ the points where the boundaries of D ′ and D intersect.The line ( χ , χ ) must cross the two other disks. Without loss of generality letus assume that ( χ , χ ) is horizontal, and that starting from D it intersects first D and then D . We also assume that χ is the intersection that is the closestto D . Suppose by contradiction that D ′′ does not intersect with D ′ . let p ′′ bein D ′′ ∩ D and p ′′ be in D ′′ ∩ D . let also p ′ be in D ′ ∩ D . Note that p ′ isbelow the line ( χ , χ ) . As all the part of D that is below ( χ , χ ) is containedin D ′ , therefore the line segment [ p ′′ , p ′′ ] must be above the line segment [ χ , p ′ ] .Hence the line segment [ p ′′ , p ′′ ] intersect with D . As D ′′ does not contain p ′ ,it must contain D \ A . Moreover p ′′ and p ′′ are not on the same side of D ,which implies that D ′′ is centred. This is a contradiction, therefore D ′ and D ′′ intersect. Lemma 17.
If for each disk D i there exists a line transversal with disk in themiddle being D i , then it is possible to split {D ′ j } into two sets X and X , suchthat G ( X ) and G ( X ) are complete graphs.Proof. If there is a disk contained as in Lemma 12, then it is done. Otherwise weknow with Lemma 13 that each of the three disks D , D and D is -intersecting.Note that for each disk D i we have D i \ A i = ∅ . We can assume that there is aximum clique in geometric graphs 21 no disk in {D ′ j } that contains one of D , D or D . Indeed such a disk wouldintersect with all the other disks in {D ′ j } , so we could add it arbitrarily to X or X .We separate the disks in {D ′ j } into two subsets. Let X the set of all disks D ′ in {D ′ j } that are centred. Let X be defined as {D ′ j } \ X . Using Lemma 15,we immediately obtain that G ( X ) is a complete graph. Let D ′ and D ′′ be twodisks in X . If one of them is outside-containing, then it is not centred, and wecan apply Lemma 16 to show that D ′ and D ′′ intersect.From now on we assume that D ′ and D ′′ are not outside-containing. Let usassume that there exists one disk D i such that D ′ contains a point p ′ i ∈ A i , and D ′′ contains a point p ′′ i ∈ A i . Without loss of generality let us assume that thisdisk is D . Let us also assume that τ is horizontal, and that D and D are belowit. Let p ′ (respectively p ′′ ) be a point in D ′ ∩ D (respectively D ′′ ∩ D ), andlikewise for p ′ . Let us consider the triangles p ′ p ′ p ′ and p ′′ p ′′ p ′′ . We denote by χ ′ and χ ′ the points (potentially identical) where p ′ p ′ p ′ intersects the boundaryof D . All these points are in A because of it is bounded from above by τ . Wetake these points such that the x -coordinate of χ ′ is not less that the one of χ ′ .We do likewise for χ ′′ and χ ′′ . Let us examine in which order they appear onthe circular arc bounding A . Without loss of generality, let us assume that χ ′ appears first. If they appear in the following order χ ′ → χ ′′ → χ ′ → χ ′′ thenthe triangles intersect by convexity of A . If the order is χ ′ → χ ′ → χ ′′ → χ ′′ then the triangles intersect because D and D are not intersecting. Finally, inthey appear in this order χ ′ → χ ′′ → χ ′′ → χ ′ , then the triangles intersectbecause D \ A is not contained in D ′ . Indeed two circles intersect at mosttwice, therefore D ′ would contain all the surface in D that is above [ χ ′ , χ ′ ] ,which is impossible by definition of X .Now let us assume that D ′ (respectively D ′′ ) contains a point p ′ ∈ A (re-spectively p ′′ ∈ A ). Let p ′ (respectively p ′′ ) be a point in D ′ ∩ D (respectively D ′′ ∩D ). If [ p ′ , p ′ ] and [ p ′′ , p ′′ ] intersect then it is done. Otherwise let us assumewithout loss of generality that [ p ′ , p ′ ] is above [ p ′′ , p ′′ ] . If p ′′ is in A then we canapply the previous reasoning to p ′ and p ′′ . Otherwise the line segment [ p ′′ , p ′′ ] crosses τ . Therefore the line segment [ p ′ , p ′ ] must cross [ p ′ , p ′′ ] . Note that isdoes not cross τ , because p ′ and p ′ are on the same side of it. Now as D ′ doesnot contain D \ A , it must intersect with D ′′ .Let us assume that D ′ contains a point p ′ ∈ A , and that for i = 1 , , , wehave D ′′ ∩ A i = ∅ . Let p ′′ be a point in D ′′ ∩ D , and p ′′ in D ′′ ∩ D . By convexityof D ′′ , the line segment [ p ′′ , p ′′ ] does not intersect with A or A . Without lossof generality, let us assume that D is on the left hand side. If the line segment [ p ′′ , p ′′ ] is on the right hand side of A , then it intersect with [ p ′ , p ′ ] , for any p ′ in D ′ ∩ D . If the line segment [ p ′′ , p ′′ ] is on the left hand side of A , then [ p ′′ , p ′′ ] must be below A . Therefore [ p ′′ , p ′′ ] intersect with [ p ′ , p ′ ] . In any case, the twodisks D ′ and D ′′ intersect.Finally let us assume that for i = 1 , , , we have D ′′ ∩ A i = D ′′ ∩ A i = ∅ . Letus consider [ p ′ , p ′ ] and [ p ′′ , p ′′ ] . If they intersect then we are done. Otherwiselet us assume without loss of generality that [ p ′ , p ′ ] is on the left hand side of [ p ′′ , p ′′ ] , and that D is on the right hand side of both. Then as [ p ′ , p ′ ] cannotcross τ , and [ p ′ , p ′ ] cannot cross τ , the two disks D ′ and D ′′ must intersect.We have shown that G ( X ) is a complete graph. Acknowledgements
The author wants to thank Michael Hoffmann for his advice and his help con-cerning the writing of the paper.
References
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A A motivation for Π d As we define it, Π d is the class of intersection graphs of d -pancakes and unit d -balls. The properties that we desire are:1. Π d contains ( d − -ball graphs and unit d -ball graphs,2. Maximum clique can be computed in polynomial time in Π and with anEPTAS in Π Let { ξ i } ≤ i ≤ d be the canonical basis of R d . Let us consider an other class ˜ Π d ,that might satisfy those properties. Definition 12.
We denote by ˜ Π d the class of intersection graphs of ( d − -ballslying on the hyperspace induced by { ξ , ξ , . . . , ξ d − } and of unit d -balls. This class might look more natural since it makes use only of balls andnot of pancakes. It contains by definition ( d − -ball graphs and unit d -ballgraphs. Moreover, as we want to be able to compute a maximum clique fast,we are looking for a "small" superclass. However, while we cannot rule out theexistence of a polynomial algorithm for computing a maximum clique on ˜ Π ,nor of an EPTAS in ˜ Π , we demonstrate that Lemma 4 does not hold in ˜ Π ,and that Conjecture 1 does not hold in ˜ Π . A.1 In R The counterexample for ˜ Π is illustrated in Figure 12. We have two intersectingunit disks D and D ′ . Moreover each D , D and the line segment [ x , x ] intersectwith both D and D ′ . The distances d ( D , D ) , d ( D ′ , D ) are smaller than d ( D , D ′ ) ,and the same hold for D . To define L ( D ) , a natural way would be to use thesame characterisation as in Observation 1. Therefore the line segment [ x , x ] isnot in L ( D ) nor in L ( D ′ ) . However, G ( {D , D , [ x , x ] } ) is an edgeless graphwith three vertices, and therefore is not cobipartite. A.2 In R We show that Conjecture 1 does not hold in ˜ Π . For any K , we give a graph G in ˜ Π with iocp ( G ) = K . With disks in the plane, Bonnet et al. have shownhow to pack arbitrarily many even cycles and one odd cycle [5]. We use here asimilar idea, but using unit balls and disks. For sake of simplicity, we only give aconstruction using odd cycles with size , but the construction could be extendedto higher cardinalities. Figure 13 shows the construction of one cycle with size .There are two disks (which happen to be unit disks) and three unit balls, whosecentres are above the plane xOy . In Figure 13 we represent only the intersectionof a unit ball with the plane xOy . For sake of clarity, we have exaggerated afew things in the representation. The centre of B is slightly above the centre of D . Therefore in reality the disk obtained as B ∩ xOy is very close to D . Thecentre of B is on the plane xOy . We rotate B and D around D so that their aximum clique in geometric graphs 25 D D ′ x x D D • • Fig. 12.
Lemma 4 does not hold in ˜ Π centres are very close. We then shift B slightly so that it does not intersect D .The height of the centre of B is very close to , therefore its intersection with xOy is very small. We also move slightly B toward B so that B ∩ xOy doesnot intersect with D nor with D . The reader can check that this is indeed arepresentation of an odd cycle of size in ˜ Π . Note that this construction couldnot be done in Π . Indeed one could think of replacing the disks by -pancakes,with the radius of the pancakes being -minus the radius of the disks. Howeveras D would be "filled" with unit balls, it would intersect B . One other ideawould be to add to the height of all unit balls, and to represent the disks asthe top of -pancakes. But then D would be intersecting with B .Now it remains to pack indefinitely many of these odd cycles. To do this, weconsider the where the centre c of B ∩ xOy was, before we moved B toward B .Then we pack the odd cycles by rotating around c . The rotation is done coun-terclockwise. When packing a second odd cycle, the new ball B ′ still intersectswith B . Their intersection is not on the plane xOy as for B and B , but theyintersect above. B Recognition of graphs in Π We show that testing whether a graph can be obtained as the intersection graphof unit disks and -pancakes is hard, as claimed in Theorem 2. Proof of Theorem 2.
We do a reduction from recognition of unit disk graphs,which is ∃ R -hard as shown by Kang and Müller [12]. Let G = ( V, E ) be a graphwith n vertices. We are going to construct (cid:0) n (cid:1) graphs such that G is a unit diskgraph if and only at least one of these new graphs is in Π . Let S and S ′ betwo stars with internal vertex s and s ′ respectively, having n + 8 leaves each.Let W and W ′ be two paths with n vertices each with end vertices w , w n and w ′ , w ′ n respectively. Let u and v be two vertices in V . We define G u,v asthe graph obtained by connecting s to s ′ , w to u , w ′ to v , w n to s and w ′ n B D B D B Fig. 13.
An odd cycle in ˜ Π to s ′ . We claim that G is a unit disk graph if and only if G u,v is in Π forsome u and v in V . First let us assume that G is a unit disk graph. Let usconsider the position of the unit disks centres in a fixed representation of G .Take two unit disks D and D ′ whose centres are in the convex hull, and let usdenote by u and v the corresponding vertices. Now take two sets {D i } ≤ i ≤ n and {D ′ j } ≤ j ≤ n of n unit disks such that G ( {D i } ) and G ( {D ′ j } ) are paths, andsuch that no two unit disks of the form D i , D ′ j intersect. Moreover we requirethat G ( {D i } ) ∩ G = ( { u } , ∅ ) and G ( {D j } ) ∩ G = ( { v } , ∅ ) , and that all unit diskscentres are on the same side of the line ( c n , c ′ n ) , which are the centres of D n and D ′ n respectively. This is possible because the most distant points in the unitdisk representation of G have distance at most n , and we have n unit disks ineach path. Then we translate and rotate everything so that the y -coordinate of c n and c ′ n is equal to , and that all other centres are above the horizontal linewith height . We take two intersecting -pancakes such that one also intersectwith D n and the other with D ′ n . We choose these -pancakes big enough sothat for each of them we can add n + 8 pairwise non intersecting unit disks,but intersecting with their respective -pancake. This shows that if G is a unitdisk graph, then G u,v is in Π .Let us now assume that G u,v is in Π , for some u, v in V . As a unit disk canintersect at most with pairwise non intersecting unit disks, we have that inany Π representation of G u,v , s and s ′ must be represented by -pancakes,denoted by P and P ′ respectively. Let x be the length of the line segmentobtained as the intersection of P and Ox . Note that all points of a unit diskintersecting a pancake is within distance of Ox . Therefore, the disks and -pancakes corresponding to the leaves of S are contained in a rectangle with areasmaller than x +4) . We have n +8 unit disks in this area. As the area of a unitdisk is bigger than , we have x + 4) ≥ n + 8) , or equivalently x ≥ n . Note aximum clique in geometric graphs 27 that the same holds with P ′ . Let us show that in any Π representation of G u,v ,all the vertices in V are represented by unit disks. Assume by contradiction thatis not. Without loss of generality, let us assume P is to the left of P ′ , and thatone vertex u G in V is represented by a -pancake that is to the right of P ′ . Indeedthis -pancake cannot be between P and P ′ because they are intersecting. Letus consider the last vertex in a path from s to u G that is a disk. By construction,the distance between P and the unit disk corresponding to this vertex is at most n + n −
1) = 6 n − . This shows that this vertex is still far from the rightend of P ′ , and so the next vertex has to be represented by a unit disk because itis not intersecting P ′ , which is a contradiction. We have shown that G is a unitdisk graph if and only if there exist u, v in V such that G u,v is in Π , and theconstruction of these (cid:0) n (cid:1) graphs takes linear time. C Proof of Theorem 3
Before stating the algorithm, we give some definitions.Vapnik and Chervonenkis have introduced the concept of VC-dimensionin [15]. In this paper, we are only concerned with the VC-dimension of the neigh-bourhood of some geometric intersection graphs. In this context, the definitioncan be stated as follow:
Definition 13.
Let F be a family of sets in R d , and let G be the intersectiongraph of F . We say that F ⊆ F is shattered if for every subset X of F , thereexists a vertex v in G that is adjacent to all vertices in X , and adjacent to novertex in F \ X . The VC-dimension of the neighbourhood of G is the maximumcardinality of a shattered subset of F . We define the class X ( d, β, K ) as introduced by Bonamy et al. in [4]. Let d and K be in N , and let β be a real number such that < β ≤ . Then X ( d, β, K ) denotes the class of simple graphs G such that the VC-dimension of the neigh-bourhood of G is at most d , α ( G ) ≥ β | V ( G ) | , and iocp ( G ) ≤ K . They showthat there exist EPTAS (Efficient Polynomial-Time Approximation Scheme) forcomputing a maximum independent set in X ( d, β, K ) . An EPTAS for a maximi-sation problem is an approximation algorithm that takes a parameter ε > andoutputs a (1 − ε ) -approximation of an optimal solution, and running in f ( ε ) n O (1) time. More formally, we have the following: Theorem 6 (Bonamy et al. [4]).
For any constants d, K ∈ N , < β ≤ , forevery ε > , there is a randomised (1 − ε ) -approximation algorithm running intime ˜ O (1 /ε ) n O (1) for maximum independent set on graphs of X ( d, β, K ) with n vertices. In this section we show the following:
Theorem 7.
If Conjecture 1 holds, then for every ε > , there is a randomised (1 − ε ) -approximation algorithm running in time ˜ O (1 /ε ) n O (1) for maximumclique on graphs of Π with n vertices. Theorem 6 states that there exists an EPTAS for computing a maximumindependent set on graphs of X ( d, β, K ) , for any d, K ∈ N and < β ≤ . Let G be in Π . In order to prove Theorem 7, we show that the VC-dimension of theneighbourhood of any vertex in G is bounded. Remark that the VC-dimension ofa graph and its complement are equal. We aim at using the EPTAS mentionedabove for computing a maximum independent set in the complement, which isequivalent to computing a maximum clique in the original graph. However agraph G in Π does not necessarily satisfy α ( G ) ≥ β | V ( G ) | for some < β ≤ .Even if it does, we need to know the value of β in order to use the EPTASof Theorem 6. Therefore we show how to compute a maximum clique in any G ∈ Π by using polynomially many times the EPTAS of Theorem 6 on somesubgraphs of G , which have the desired property.In general, for intersection graphs of geometric objects that can be de-scribed with finitely many parameters, the VC-dimension of the neighbourhoodis bounded. For graphs in Π , we were able to show an upper bound of . Wedo not expect this value to be tight, but showing any constant was sufficient forour purpose. Proposition 1.
The VC-dimension of the neighbourhood of a graph G = ( V, E ) in Π is at most . We use the fact that the VC-dimension of the neighbourhood of disk graphs(and even pseudo-disk graphs) is at most , as proved by Aronov et al. [1].Likewise, the VC-dimension of the neighbourhood of unit ball graphs is at most , as noticed by Bonamy et al. [4]. Proof of Proposition 1.
First let us show that if V is shattered, then in any Π representation of G there are at most four -pancakes. Let us assume bycontradiction that there exists a set S of five -pancakes, such that for everysubset T of S , there exists a unit ball or a -pancake intersecting the set X exactly on T . For each -pancake P ( c i , ρ ) in S , we denote by D i the disk D ( c i , ρ ) lying on the plane xOy . Let T be a subset of S . If there exists a -pancake P ( c ′ , ρ ′ ) intersecting exactly with the elements of T , we denote by D T the disk D ( c ′ , ρ ′ + 2) lying on the plane xOy . Otherwise there exists a unit ball B centred at c ′′ intersecting exactly with the elements of T , and then we denote by D T the intersection between B ( c ′′ , and xOy . As B intersects with a -pancake, D T is not empty. Remark that in both cases D i intersects with D T if and onlyif P ( c i , ρ ) is in T . This implies that if S is shattered by some -pancakes andunit balls, then the set {D i } is shattered by {D T | T ⊆ S } . However this is notpossible because the VC-dimension of the neighbourhood of disk graphs is atmost .Now let us prove the claim. Assume by contradiction that we have a shatteredset with elements. As shown above, in any Π representation there are at least unit balls. Let us consider such a representation, we denote by S , . . . , S fivesets of five unit balls each. As the VC-dimension of the neighbourhood of unitball graphs is at most , for each set S i there exists a non-empty subset T i ⊆ S i such that no unit ball can intersect with the elements in T i , but not with those in aximum clique in geometric graphs 29 S i \ T i . For each T i , we choose arbitrarily one element t i , and define a new set T that contains the five t i . Moreover for each unit ball B centred at c correspondingto a t i , we denote by D i the intersection between B ( c, and the plane xOy . Let T ′ be a subset of T , and let us consider the set ∪ t i ∈ T ′ T i , that we denote by T ′ + .Notice that unless T ′ = ∅ , no unit ball can intersect with all elements in T ′ + andwith no element in S \ T ′ + . Therefore this can only be achieved by a -pancake,and we denote by D T its intersection with the plane xOy . Therefore the fivedisks D i are shattered by disks, which is impossible. Proof of Theorem 7.