On disjoint crossing families in geometric graphs
OOn disjoint crossing families in geometric graphs
Radoslav Fulek ∗† Andrew Suk ∗‡ November 10, 2018
Abstract A geometric graph is a graph drawn in the plane with vertices represented by points andedges as straight-line segments. A geometric graph contains a ( k, l ) -crossing family if there isa pair of edge subsets E , E such that | E | = k and | E | = l , the edges in E are pairwisecrossing, the edges in E are pairwise crossing, and every edges in E is disjoint to every edge in E . We conjecture that for any fixed k, l , every n -vertex geometric graph with no ( k, l )-crossingfamily has at most c k,l n edges, where c k,l is a constant that depends only on k and l . In thisnote, we show that every n -vertex geometric graph with no ( k, k )-crossing family has at most c k n log n edges, where c k is a constant that depends only on k , by proving a more general resultwhich relates extremal function of a geometric graph F with extremal function of two completelydisjoint copies of F . We also settle the conjecture for geometric graphs with no (2 , F on 3 vertices, every n -vertex geometric graph that does not contain a matching whose intersection graph is F hasat most O ( n ) edges. A topological graph is a graph drawn in the plane with points as vertices and edges as non-self-intersecting arcs connecting its vertices. The arcs are allowed to intersect, but they may notpass through vertices except for their endpoints. Furthermore, the edges are not allowed to havetangencies, i.e., if two edges share an interior point, then they must properly cross at that point.We only consider graphs without parallel edges or self-loops. A topological graph is simple if everypair of its edges intersect at most once. If the edges are drawn as straight-line segments, then thegraph is geometric . Two edges of a topological graph cross if their interiors share a point, and are disjoint if they do not have a point in common (including their endpoints).It follows from Euler’s Polyhedral Formula that every simple topological graph on n verticesand no crossing edges has at most 3 n − n vertices with no pair of disjoint edges has at most O ( n ) edges [10],[7]. Finding themaximum number of edges in a topological (and geometric) graph with a forbidden substructurehas been a classic problem in extremal topological graph theory (see [1], [2], [15], [6], [20], [14],[19], [18], [21]). Many of these problems ask for the maximum number of edges in a topological (or ∗ The authors gratefully acknowledge support from the Swiss National Science Foundation Grant No. 200021-125287/1 † EPFL, Lausanne. Email: [email protected] ‡ Courant Institute, New York and EPFL, Lausanne. Email: [email protected] a r X i v : . [ m a t h . C O ] M a r eometric) graph whose edge set does not contain a matching that defines a particular intersectiongraph. Recall that the intersection graph of objects C in the plane is a graph with vertex set C ,and two vertices are adjacent if their corresponding objects intersect. Much research has beendevoted to understanding the clique and independence number of intersection graphs due to theirapplications in VLSI design [8], map labeling [3], and elsewhere.Recently, Ackerman et al. [4] defined a natural ( k, l ) -grid to be a set of k pairwise disjoint edgesthat all cross another set of l pairwise disjoint edges. They conjectured Conjecture 1.1.
Given fixed constants k, l ≥ there exists another constant c k,l , such that anygeometric graph on n vertices with no natural ( k, l ) -grid has at most c k,l n edges. They were able to show,
Theorem 1.2. [4]
For fixed k , an n -vertex geometric graph with no natural ( k, k ) -grid has at most O ( n log n ) edges. Theorem 1.3. [4] An n -vertex geometric graph with no natural (2 , -grid has at most O ( n ) edges. Theorem 1.4. [4] An n -vertex simple topological graph with no natural ( k, k ) -grid has at most O ( n log k − n ) edges. As a dual version of the natural ( k, l )-grid, we define a ( k, l ) -crossing family to be a pair of edgesubsets E , E such that1. | E | = k and | E | = l ,2. the edges in E are pairwise crossing,3. the edges in E are pairwise crossing,4. every edge in E is disjoint to every edge in E .We conjecture Conjecture 1.5.
Given fixed constants k, l ≥ there exists another constant c k,l , such that anygeometric graph on n vertices with no ( k, l ) -crossing family has at most c k,l n edges. It is not even known if all n -vertex geometric graphs with no k pairwise crossing edges has O ( n )edges. The best known bound is due to Valtr [22], who showed that this is at most O ( n log n ) forevery fixed k . We extend this result to ( k, k )-crossing families by proving the following theorem. Theorem 1.6. An n -vertex geometric graph with no ( k, k ) -crossing family has at most c k n log n edges, where c k is a constant that depends only on k . Let F denote a geometric graph. We say that a geometric graph G contains F as a geometricsubgraph if G contains a subgraph F (cid:48) isomorphic to F such that two edges in F (cid:48) cross if and onlyif the two corresponding edges cross in F .We define ex ( F, n ) to be the extremal function of F , i.e. the maximum number of edges ageometric graph on n vertices can have without containing F as a geometric subgraph. Similarly,we define ex L ( F, n ) to be the extremal function of F , if we restrict ourselves to the geometric graphsall of whose edges can be hit by one line.Let F denote a geometric graph, which consists of two completely disjoint copies of a geometricgraph F . We prove Theorem 1.6 by a straightforward application of the following result.2 a) 3 pairwise crossing. (b) 3 pairwisedisjoint. (c) (2,1)-grid. (d) (2,1)-crossingfamily. Figure 1: Triples of segments corresponding to all circle graphs on three vertices.
Theorem 1.7. ex ( F , n ) = O (( ex L ( F, n ) + n ) log n + ex ( F, n ))Furthermore, we settle Conjecture 1.5 in the first nontrivial case.
Theorem 1.8. An n -vertex geometric graph with no (2 , -crossing family has at most O ( n ) edges. Note that Conjecture 1.5 is not true for topological graphs since Pach and T´oth [16] showed thatthe complete graph can be drawn such that every pair of edges intersect once or twice.Recall that F is a circle graph if F can be represented as the intersection graph of chords on acircle. By combining Theorem 1.8 with results from [2], [4], and [20], we have the following. Corollary 1.9.
For any circle graph F on vertices, every n -vertex geometric graph that does notcontain a matching whose intersection graph is F contains at most O ( n ) edges. See Figure 1. We also conjecture the following.
Conjecture 1.10.
For any circle graph F on k vertices, there exists a constant c k such that every n -vertex geometric graph that does not contain a matching whose intersection graph is F , containsat most c k n edges. As pointed out by Klazar and Marcus [9], it is not hard to modify the proof of the Marcus-TardosTheorem [19] to show that Conjecture 1.10 is true when the vertices are in convex position.For simple topological graphs, we have the following
Theorem 1.11. An n -vertex simple topological graph with no ( k, -crossing family has at most n (log n ) O (log k ) edges. The paper is organized as follows. Section 2 is devoted to the proof of Theorem 1.7. In Section3 we establish Theorem 1.8. Finally, the result of Theorem 1.11 about topological graphs is provedin Section 4.
First, we prove a variant of Theorem 1.7 when all of the edges in our geometric graph can be hitby a line. As in the introduction let F denote a geometric graph, which consists of two completelydisjoint copies of a geometric graph F . We will now show that the extremal function ex L ( F , n ) isnot far from ex L ( F, n ). 3 heorem 2.1. ex L ( F , n ) ≤ O (( n + ex L ( F, n )) log n ) . Proof.
Let G denote a geometric graph on n vertices that does not contain F as a geometricsubgraph, and all the edges of G can be hit by a line. By a standard perturbation argument wecan assume that the vertices of G are in general position. As in [5], a halving edge uv is a pair ofthe vertices in G such that the number of vertices on each side of the line through u and v is thesame. Lemma 2.2.
There exists a directed line (cid:126)l such that the number of edges in G that lies completelyto the left or right of (cid:126)l is at most ex L ( F, n/
2) + 5 n .Proof. If n is odd we can discard one vertex of G , thereby loosing at most n edges. Therefore wecan assume n is even, and it suffices to show that there exists a directed line (cid:126)l such that the numberof edges in G that lies completely to the left or right of (cid:126)l is at most 2 ex L ( F, n/
2) + 4 n .Let uv be a halving edge, and let (cid:126)l denote the directed line containing vertices u and v withdirection from u to v . Let e ( (cid:126)l, L ) and e ( (cid:126)l, R ) denote the number of edges on the left and right sideof (cid:126)l respectively. Without loss of generality, we can assume that e ( (cid:126)l, L ) ≤ e ( (cid:126)l, R ). We will rotate (cid:126)l such that it remains a halving line at the end of each step, until it reaches a position where thenumber of edges on both sides of (cid:126)l is roughly the same.We start by rotating (cid:126)l counterclockwise around u until it meets the next vertex w of G . Ifinitially w lies to the right of (cid:126)l , then in the next step we will rotate (cid:126)l around u (again). See Figure2(a). Otherwise if w was on the left side of (cid:126)l , then in the next step we will rotate (cid:126)l around vertex w . See Figure 2(b). Clearly after each step in the rotation, there are exactly n/ (cid:126)l . u v −→ lw (a) w lies to the rightof (cid:126)l . u v −→ lw (b) w lies to the left of (cid:126)l . Figure 2: Halving the vertices of G After several rotations, (cid:126)l will eventually contain points u and v again, with direction from v to u . At this point we have e ( (cid:126)l, L ) ≥ e ( (cid:126)l, R ). Since the number of edges on the right side (and leftside) changes by at most n after each step in the rotation, at some point in the rotation we musthave | e ( (cid:126)l, L ) − e ( (cid:126)l, R ) | ≤ n. Since G does not contain a F as a geometric subgraph, this implies that4 l ′ V V V ( l ′ ) V ( l ′ ) V ( l ′ ) V ( l ′ ) l ′ Figure 3: The final partition of the vertex set of Ge ( (cid:126)l, L ) ≤ ex L ( F, n/
2) + 2 n and e ( (cid:126)l, R ) ≤ ex L ( F, n/
2) + 2 n. Therefore for any n , there exists a directed line (cid:126)l such that the number of edges in G that liescompletely to the left or right of (cid:126)l is at most 2 ex L ( F, n/
2) + 5 n .By Lemma 2.2 we obtain a line l , which partition the vertices of G into two equal (or almostequal if n is odd) sets V and V . Let E (cid:48) denote the set of edges between V and V . By theHam-Sandwich Cut Theorem [11], there exists a line l (cid:48) that simultaneously bisects V and V . Let V ( l (cid:48) ) and V ( l (cid:48) ) denote the resulting parts of V , and let V ( l (cid:48) ) and V ( l (cid:48) ) denote the resultingparts of V .Observe that we can translate l (cid:48) along l into a position where the number of edges in E (cid:48) thatlie completely to the left and completely to the right of l (cid:48) is roughly the same. In particular, wecan translate l (cid:48) along l such that the number of edges in E (cid:48) that lies completely to its left or rightside is at most ex L ( F, n ) + ex L ( F, n/ n (see Figure 2). Indeed, assume that the number ofedges in E (cid:48) between, say, V ( l (cid:48) ) and V ( l (cid:48) ) is more than ex L ( F, n/ l (cid:48) to theright, the number of edges that lie completely to the right of l (cid:48) changes by at most n as l (cid:48) crossesa single vertex in G . Therefore we can translate l (cid:48) into the leftmost position where the numberof edges in E (cid:48) between V ( l (cid:48) ) and V ( l (cid:48) ) drops below ex L ( F, n/ n + 1. Moreover, at thisposition the number of edges in E (cid:48) between V ( l (cid:48) ) and V ( l (cid:48) ) still cannot be more than ex L ( F, n )since G does not contain F as a geometric subgraph.Thus, all but at most 3 ex L ( F, n/ ex L ( F, n ) + 6 n edges of G are the edges between V ( l (cid:48) )and V ( l (cid:48) ), and between V ( l (cid:48) ) and V ( l (cid:48) ). Notice that there exists k , − / ≤ k ≤ /
4, such that | V ( l (cid:48) ) | + | V ( l (cid:48) ) | = n (1 / k ), and | V ( l (cid:48) ) | + | V ( l (cid:48) ) | = n (1 / − k ). Finally, we are in the positionto state the recurrence, whose closed form gives the statement of the theorem. ex L ( F , n ) ≤ ex L ( F , n (1 / k )) + ex L ( F , n (1 / − k )) + 3 ex L ( F, n/ ex L ( F, n ) + 6 n.
5y a routine calculation which is indicated below, we have ex L ( F , n ) ≤ log (cid:18) n (cid:18)
12 + k (cid:19)(cid:19) (cid:18) n (cid:18)
12 + k (cid:19) + 4 ex L (cid:18) F , n (cid:18)
12 + k (cid:19)(cid:19)(cid:19) +log (cid:18) n (cid:18) − k (cid:19)(cid:19) (cid:18) n (cid:18) − k (cid:19) + 4 ex L (cid:18) F , n (cid:18) − k (cid:19)(cid:19)(cid:19) +4 ex L ( F, n ) + 6 n ≤ log n (4 ex L ( F, n ) + 6 n )Finally, we show how Theorem 2.1 implies Theorem 1.7. Proof of Theorem 1.7.
Let G = ( V, E ) denote the geometric graph not containing F as a subgraph.Similarly, as in the proof of Lemma 2.2 we can find a halving line l that hits all but 2 ex ( F, n/ n edges of G . Now, the claim follows by using Theorem 2.1.Theorem 1.6 follows easily by using Theorem 1.7 with a result from [22], which states thatevery n -vertex geometric graph whose edges can be all hit by a line and does not contain k pairwisecrossing edges has at most O ( n ) edges and at most O ( n log n ) edges if we do not require a singleline to hit all the edges. In this section we will prove Theorem 1.8. Our main tool is the following theorem by T´oth andValtr
Theorem 3.1. [21]
Let G = ( V, E ) be an n -vertex geometric graph. If G does not contain amatching consisting of pairwise disjoint edges, then | E ( G ) | ≤ n + 64 . Theorem 1.8 immediately follows from the following theorem.
Theorem 3.2.
Every n -vertex geometric graph with no (2 , -crossing family has at most n + 64 edges.Proof. For sake of contradiction, let G = ( V, E ) be a vertex-minimal counter example, i.e. G isa geometric graph on n vertices which has more than 64 n + 64 edges and G does not contain a(2,1)-crossing family. Hence every vertex in G has degree at least 65. Let M denote the maximummatching in G consisting of pairwise disjoint edges and let V M denote the vertices in M . Since | E ( G ) | > n + 64, Theorem 3.1 implies that | M | ≥
5. We say that two edges intersect if theycross or share an endpoint. The following simple observation is crucial in the subsequent analysis.(*) An edge e ∈ E that crosses an edge of M must intersect every edge in M .Indeed, otherwise we would obtain a (2,1)-crossing family. We call an endpoint v of an edge in M good if every ray starting at v misses at least one edge in M . See Figure 4(b).6 emma 3.3. For | M | ≥ , at least | M | − of the endpoints in M are good.Proof. We proceed by induction on | M | . Assume | M | = 4. If every triple in M has a goodvertex, then clearly we have at least two good vertices. Otherwise the only matching consistingof three pairwise disjoint edges with no good vertices is the one in Figure 4(a). By a simple caseanalysis, adding a disjoint edge to this matching creates two good vertices (see Appendix A). Forthe inductive step when | M | >
4, we choose an arbitrary 4-tuple of edges in the matching. By theabove discussion, the 4-tuple has at least one good endpoint. By removing the edge incident tothis good vertex, the statement follows by the induction hypothesis. (a) (cid:0)(cid:0)(cid:1)(cid:1)(cid:0)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:1)(cid:1) (cid:0)(cid:0)(cid:1)(cid:1) (cid:0)(cid:1) v (b) Figure 4: (a) Special case in Lemma 3.3, (b) Matching M of size 3 with one good vertex v Thus by (*), a good endpoint cannot be incident to an edge that crosses any of the edges in M .Let1. V g ⊆ V M denote the set of good endpoints in M .2. V ⊂ V \ V M be the subset of the vertices such that for v ∈ V , every edge incident to v doesnot cross any of the edges in M ,3. and V = V \ ( V M ∪ V ). Hence for v ∈ V , there exists an edge incident to v that intersectsevery edge in M .See Figure 5(a). By Lemma 3.3, | V g | ≥ | M | − | V M | / −
2. By maximality of M , there are noedges between V and V and V is an independent set. Now notice the following observation. Observation 3.4.
There exists a good vertex in V g that has at least three neighbors in V .Proof. For sake of contradiction, suppose that each vertex in V g has at most two neighbors in V .Then let G (cid:48) = ( V (cid:48) , E (cid:48) ) denote a subgraph of G such that V (cid:48) = V M ∪ V and E (cid:48) consists of theedges that do not cross any of the edges of M and whose endpoints are in V ∪ V M . Since | M | ≥ G (cid:48) must be a planar graph since otherwise we would have a (2 , E (cid:48) ≤ | V | + | V M | ). 7n the other hand, by minimality of G , each vertex in V has degree at least 65 in G (cid:48) , and eachvertex in V g has degree at least 63 in G (cid:48) . Therefore by applying Lemma 3.3, we have12 (cid:18) | V | + 63 (cid:18) | V M | − (cid:19)(cid:19) ≤ | E (cid:48) | ≤ | V | + 3 | V M | . This implies 59 | V | + 25 | V M | ≤ | V M | ≥
10 ( | M | ≥ (cid:3) Let v ∈ V g be a good vertex such that v has at least 3 neighbors in V . Let e = vv , f = vv , g = vv , and m be edges in G such that v , v , v ∈ V , m ∈ M , and v is a good vertex incidentto m . Furthermore, we will assume that g, e, m, f appear in clockwise order around v . By (*)there is an edge e (cid:48) incident to v having non-empty intersection with every edge in M . Similarly,we can find such an edge f (cid:48) incident to v (possibly f (cid:48) = e (cid:48) ). The edges e (cid:48) and f (cid:48) must havenon-empty intersection with f and e , respectively (see Figure 5(b)). Otherwise we would obtain a(2,1)-crossing family in G consisting of e, f (cid:48) and an edge from M , or e (cid:48) , f and an edge from M .However, a third edge g cannot have a non-empty intersection with both e (cid:48) and f (cid:48) . Hence, weobtain a (2,1)-crossing family in G consisting of g, e (cid:48) and an edge from M , or g, f (cid:48) and an edgefrom M . Thus, there is no minimal counter example and that concludes the proof. (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) V V M (a) v v v e ′ f ′ g (b) Figure 5: (a) M , V , and V , (b) Situation around the vertex v We note that by a more tedious case analysis, one could improve the upper bound in Theorem 3.2to 15 n . ( k, -crossing family In this section, we will prove Theorem 1.11 which will require the following two lemmas. The firstone is due to Fox and Pach.
Lemma 4.1. [6]
Every n -vertex simple topological graph with no k pairwise crossing edges has atmost n (log n ) c log k edges, where c is an absolute constant.
8s defined in [17], the odd-crossing number odd-cr( G ) of a graph G is the minimum possiblenumber of unordered pairs of edges that crosses an odd number of times over all drawings of G . The bisection width of a graph G , denoted by b ( G ), is the smallest nonnegative integer suchthat there is a partition of the vertex set V = V ˙ ∪ V with · | V | ≤ V i ≤ · | V | for i = 1 , | E ( V , V ) | = b ( G ). The second lemma required is due to Pach and T´oth, which relates theodd-crossing number of a graph to its bisection width. Lemma 4.2. [16]
There is an absolute constant c such that if G is a graph with n vertices ofdegrees d , . . . , d n , then b ( G ) ≤ c log n (cid:118)(cid:117)(cid:117)(cid:116) odd-cr( G ) + n (cid:88) i =1 d i . Since all graphs have a bipartite subgraph with at least half of its edges, Theorem 1.6 immediatelyfollows from the following Theorem.
Theorem 4.3.
Every n vertex simple topological bipartite graph with no ( k, -crossing family hasat most c n log c log k n edges, where c , c are absolute constants.Proof. We proceed by induction on n . The base case is trivial. For the inductive step, the prooffalls into two cases. Case 1.
Suppose there are at least | E ( G ) | / ((2 c ) log n ) disjoint pair of edges in G . Then bydefining D ( e ) to be the set of edges disjoint from edge e , we have2 | E ( G ) | (2 c ) log n ≤ (cid:80) e ∈ E ( G ) | D ( e ) || E ( G ) | Hence there exists an edge that is disjoint to at least 2 | E ( G ) | / ((2 c ) log n ) other edges. ByLemma 4.1 we have 2 | E ( G ) | (2 c ) log n ≤ n (log n ) c log k , which implies | E ( G ) | ≤ c n log c log k n for sufficiently large constants c , c . Case 2.
Suppose there are at most | E ( G ) | / ((2 c ) log n ) disjoint pair of edges in G . Since G isbipartite, let V a and V b be its vertex class. By applying a suitable homeomorphism to the plane,we can redraw G such that1. the vertices in V a are above the line y = 1, the vertices in V b are below the line y = 0,2. edges in the strip 0 ≤ y ≤ G that lies above the y = 1 line about the y -axis. Then erase the edgesin the strip 0 ≤ y ≤ (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:1)(cid:1)(cid:0)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:1) (cid:0)(cid:0)(cid:1)(cid:1)(cid:0)(cid:1) (cid:0)(cid:1)(cid:0)(cid:1) (cid:0)(cid:0)(cid:1)(cid:1) (cid:0)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1) y=1y=0 Figure 6: Redrawing procedurepairs on the line y = 0 and y = 1. See Figure 4, and note that our graph is no longer simple. Sincethere are at most (cid:80) v ∈ V ( G ) d ( v ) ≤ | E ( G ) | n pair of edges that share a vertex in G , this impliesodd-cr( G ) ≤ | E ( G ) | (2 c ) log n + 2 | E ( G ) | n. By Lemma 4.2, there is a partition of the vertex set V = V ˙ ∪ V with · | V | ≤ V i ≤ · | V | for i = 1 , b ( G ) ≤ c log n (cid:115) | E ( G ) | (2 c ) log n + 4 n | E ( G ) | . If | E ( G ) | (2 c ) log n ≤ n | E ( G ) | then we have | E ( G ) | ≤ c n log c log k n and we are done. Therefore we can assume b ( G ) ≤ c log n (cid:115) | E ( G ) | (2 c ) log n ≤ | E ( G ) | log n . Let | V | = n and | V | = n . By the induction hypothesis we have | E ( G ) | ≤ b ( G ) + c n log c log k n + c n log c log k n ≤ | E ( G ) | log n + c n log c log k (2 n/ ≤ | E ( G ) | log n + c n (log n − log(3 / c log k , which implies | E ( G ) | ≤ c n log c log k n (1 − log(3 / / log n ) c log k − / log n ≤ c n log c log k n. For small values of k , one can obtain better bounds by replacing Lemma 4.1 with a Theorem ofPach et. al. [14] and Ackerman [1] to obtain Theorem 4.4.
For k > , every n vertex simple topological graph with no ( k, -crossing familyhas at most O (cid:0) n log k +2 n (cid:1) edges. For k = 2 , , , every n vertex simple topological graph with no ( k, -crossing family has at most O (cid:0) n log n (cid:1) edges. (cid:3) Appendix
A Four disjoint edges (a) (b) (c) (d) (e)(f) (g)
Figure 7: Possible configuration of 4 pairwise disjoint edges, the additional edge is bold, and goodvertices are marked by small discsSuppose that we have three pairwise disjoint edges in the plane, whose combinatorial configura-tion is that of the configuration in Figure 4(a). Let T denote the triangle we get by prolonging theedges until they hit another edge. In what follows we show that by adding additional edge to thisconfiguration, so that all the edges remain pairwise disjoint, we obtain at least two good endpoints(as defined in the proof of Theorem 1.8).There are three cases to check:1. The additional edge is completely inside of the triangle T (see Figure 7(a)). The two goodpoints are the endpoints of the additional edge.2. The additional edge e has one endpoint in the inside the triangle T and the one outside ofit. Clearly, the endpoint of e inside of T is a good endpoint. If the other endpoint of e is notgood, we can easily see (see Figure 7(b), 7(c) and 7(d)), that one of the remaining endpointsis good. 11. The additional edge is completely outside of the triangle T . There are three cases to bechecked (see Figure 7(e), 7(f) and 7(g)), according to where the lines through our segmentsmeet. References [1] E. Ackerman, 2006. On the maximum number of edges in topological graphs with no four pair-wise crossing edges. In Proceedings of the Twenty-Second Annual Symposium on ComputationalGeometry (Sedona, Arizona, USA, June 05 - 07, 2006). SCG ’06. ACM, New York, NY, 259-263.[2] P.K. Agarwal, B. Aronov, J. Pach, R. Pollack, and M. Sharir, Quasi-Planar Graphs Have aLinear Number of Edges. In Proceedings of the Symposium on Graph Drawing (September 20 -22, 1995). F. Brandenburg, Ed. Lecture Notes In Computer Science, vol. 1027. Springer-Verlag,London, 1-7.[3] P.K. Agarwal, M. van Kreveld, and S. Suri, Label placement by maximum independent set inrectangles,
Comput. Geom. Theory Appl. (1998), 209–218.[4] E. Ackerman, J. Fox, J. Pach, and A. Suk, On grids in topological graphs, Proc. 25th ACMSymp. on Computational Geometry (SoCG), University of Aarhus, Denmark, June 2009, 403-412.[5] T. Dey, Improved Bounds on Planar k-sets and k-levels, Discrete Comput. Geom, 1997, 19,156–161[6] J. Fox and J. Pach, Coloring K k -free intersection graphs of geometric objects in the plane.In Proceedings of the Twenty-Fourth Annual Symposium on Computational Geometry (CollegePark, MD, USA, June 09 - 11, 2008). SCG ’08. ACM, New York, NY, 346-354.[7] R. Fulek and J. Pach, A computational approach to Conway’s thrackle conjecture, Computa-tional Geometry: Theory and Application, 2010, to appear.[8] D. S. Hochbaum and W. Maass, Approximation schemes for covering and packing problems inimage processing and VLSI, J. ACM (1985), 130–136.[9] M. Klazar and A. Marcus,Extensions of the linear bound in the Fredi-Hajnal conjecture, Adv.in Appl. Math. 38 (2006), 258–266.[10] L. Lov´asz, J. Pach and M. Szegedy: On Conway’s thrackle conjecture, Discrete. Comput.Geom. 18(4) (1997), 369376.[11] J. Matou´sek, Lectures on Discrete Geometry. Springer-Verlag New York, Inc. 2002.[12] A. Marcus, G. Tardos, Excluded permutation matrices and the Stanley-Wilf conjecture, Jour-nal of Combinatorial Theory Series A, v.107 n.1, p.153-160, July 2004.[13] J. Pach and J. Solymosi, Crossing patterns of segments. J. Comb. Theory Ser. A 96, 2 (Nov.2001), 316-325. 1214] J. Pach, R. Radoicic, G. Toth. Relaxing planarity for topological graphs, J. Akiyama, M.Kano (Eds.): Discrete and Computational Geometry, Japanese Conference, JCDCG 2002, Tokyo,Japan, Lecture Notes in Computer Science, 2866, Springer, 2003, 221–232.[15] J. Pach, F. Shahrokhi, and M. Szegedy. 1994. Applications of the crossing number. In Pro-ceedings of the Tenth Annual Symposium on Computational Geometry (Stony Brook, New York,United States, June 06 - 08, 1994). SCG ’94. ACM, New York, NY, 198-202.[16] J. Pach and G. T´oth, Disjoint edges in topological graphs,in: Combinatorial Geometry andGraph Theory (J. Akiyama et al., eds.),Lecture Notes in Computer Science 3330,Springer-Verlag,Berlin, 2005, 133–140.[17] J. Pach and G. T´oth, Which crossing number is it anyway?. J. Comb. Theory Ser. B 80, 2(Nov. 2000), 225-246.[18] J. Pach, J. T¨or¨ocsik, Some Geometric Applications of Dilworth’s Theorem. Discrete & Com-putational Geometry 12: 1-7 (1994)[19] G. Tardos, G. T´oth: Crossing stars in topological graphs, Japan Conference on Discrete andComputational Geometry 2004, Lecture Notes in Computer Science 3742 Springer-Verlag, Berlin,184-197.[20] G. T´oth, 2000. Note on geometric graphs. J. Comb. Theory Ser. A 89, 1 (Jan. 2000), 126-132.[21] G. T´oth and P. Valtr. 1998 Geometric Graphs with Few Disjoint Edges. Technical Report.UMI Order Number: 98-22., Center for Discrete Mathematics & Theoretical Computer Science.[22] P. Valtr, Graph drawings with no kk