The number of edges in k-quasi-planar graphs
aa r X i v : . [ m a t h . C O ] D ec The number of edges in k -quasi-planar graphs ∗ Jacob Fox † J´anos Pach ‡ Andrew Suk § November 6, 2018
Abstract
A graph drawn in the plane is called k -quasi-planar if it does not contain k pairwise crossingedges. It has been conjectured for a long time that for every fixed k , the maximum numberof edges of a k -quasi-planar graph with n vertices is O ( n ). The best known upper bound is n (log n ) O (log k ) . In the present note, we improve this bound to ( n log n )2 α ck ( n ) in the specialcase where the graph is drawn in such a way that every pair of edges meet at most once. Here α ( n ) denotes the (extremely slowly growing) inverse of the Ackermann function. We also makefurther progress on the conjecture for k -quasi-planar graphs in which every edge is drawn asan x -monotone curve. Extending some ideas of Valtr, we prove that the maximum number ofedges of such graphs is at most 2 ck n log n . A topological graph is a graph drawn in the plane such that its vertices are represented by pointsand its edges are represented by non-self-intersecting arcs connecting the corresponding points. Innotation and terminology, we make no distinction between the vertices and edges of a graph andthe points and arcs representing them, respectively. No edge is allowed to pass through any pointrepresenting a vertex other than its endpoints. Any two edges can intersect only in a finite numberof points. Tangencies between the edges are not allowed. That is, if two edges share an interiorpoint, then they must properly cross at this point. A topological graph is simple if every pair ofits edges intersect at most once: at a common endpoint or at a proper crossing. If the edges of agraph are drawn as straight-line segments, then the graph is called geometric .Finding the maximum number of edges in a topological graph with a forbidden crossing patternis a fundamental problem in extremal topological graph theory (see [2, 3, 4, 6, 10, 12, 16, 21, 23]).It follows from Euler’s Polyhedral Formula that every topological graph on n vertices and with notwo crossing edges has at most 3 n − k -quasi-planar if it can be drawnas a topological graph with no k pairwise crossing edges. A graph is 2-quasi-planar if and only ∗ A preliminary version of this paper with A. Suk as its sole author will appear in
Proc. 19th Internat. Symp. onGraph Drawing (GD 2011, TU Eindhoven), LNCS , Springer, 2011 † Massachusetts Institute of Technology, Cambridge, MIT. Supported by a Simons Fellowship and NSF grant DMS1069197. Email: [email protected] ‡ EPFL, Lausanne and Courant Institute, New York, NY. Supported by Hungarian Science Foundation EuroGIGAGrant OTKA NN 102029, by Swiss National Science Foundation Grant 200021-125287/1, and by NSF Grant CCF-08-30272. Email: [email protected] . § Massachusetts Institute of Technology, Cambridge. Supported by an NSF Postdoctoral Fellowship. Email: [email protected] .
1f it is planar. According to an old conjecture (see Problem 1 in Section 9.6 of [5]), for any fixed k ≥ c k such that every k -quasi-planar graph on n vertices has at most c k n edges. Agarwal, Aronov, Pach, Pollack, and Sharir [4] were the first to prove this conjecturefor simple all k = 4.For larger values of k , first Pach, Shahrokhi, and Szegedy [18] showed that every simple k -quasi-planar graph on n vertices has at most c k n (log n ) k − edges. For k ≥ k -quasi-planar graphs, Pach, Radoiˇci´c, and T´oth [17] established the upperbound c k n (log n ) k − . Plugging into these proofs the above mentioned result of Ackerman [1], for k ≥
4, we obtain the slightly better bounds c k n (log n ) k − and c k n (log n ) k − , respectively. Forlarge values of k , the exponent of the polylogarithmic factor in these bounds was improved by Foxand Pach [10], who showed that the maximum number of edges of a k -quasi-planar graph on n vertices is n (log n ) O (log k ) .For the number of edges of geometric graphs, that is, graphs drawn by straight-line edges, Valtr[22] proved the upper bound O ( n log n ). He also extended this result to simple topological graphswhose edges are drawn as x -monotone curves [23].The aim of this paper is to improve the best known bound, n (log n ) O (log k ) , on the number ofedges of a k -quasi-planar graph in two special cases: for simple topological graphs and for notnecessarily simple topological graphs whose edges are drawn as x -monotone curves. In both cases,we improve the exponent of the polylogarithmic factor from O (log k ) to 1 + o (1). Theorem 1.1.
Let G = ( V, E ) be a k -quasi-planar simple topological graph with n vertices. Then | E ( G ) | ≤ ( n log n )2 α ( n ) ck , where α ( n ) denotes the inverse of the Ackermann function and c k is aconstant that depends only on k . Recall that the
Ackermann (more precisely, the
Ackermann-P´eter) function A ( n ) is defined asfollows. Let A ( n ) = 2 n , and A k ( n ) = A k − ( A k ( n − k = 2 , , . . . . In particular, we have A ( n ) = 2 n , and A ( n ) is an exponential tower of n two ’s. Now let A ( n ) = A n ( n ), and let α ( n ) bedefined as α ( n ) = min { k ≥ A ( k ) ≥ n } . This function grows much slower than the inverse ofany primitive recursive function. Theorem 1.2.
Let G = ( V, E ) be a k -quasi-planar (not necessarily simple) topological graph with n vertices, whose edges are drawn as x -monotone curves. Then | E ( G ) | ≤ ck n log n , where c is anabsolute constant. In both proofs, we follow the approach of Valtr [23] and apply results on generalized Davenport-Schinzel sequences. An important new ingredient of the proof of Theorem 1.1 is a corollary of aseparator theorem established in [9] and developed in [8]. Theorem 1.2 is not only more generalthan Valtr’s result, which applies only to simple topological graphs, but its proof gives a somewhatbetter upper bound: we use a result of Pettie [20], which improves the dependence on k from doubleexponential to single exponential. The sequence u = a , a , ..., a m is called l -regular if any l consecutive terms are pairwise different.For integers l, t ≥
2, the sequence S = s , s , ..., s lt
2f length lt is said to be of type up ( l, t ) if the first l terms are pairwise different and s i = s i + l = s i +2 l = · · · = s i +( t − l for every i, ≤ i ≤ l . For example, a, b, c, a, b, c, a, b, c, a, b, c, is a type up (3 ,
4) sequence or, in short, an up (3 ,
4) sequence. We need the following theorem ofKlazar [13] on generalized Davenport-Schinzel sequences.
Theorem 2.1 (Klazar) . For l ≥ and t ≥ , the length of any l -regular sequence over an n -elementalphabet that does not contain a subsequence of type up ( l, t ) has length at most n · l ( lt − · (10 l ) α ( n ) lt . For l ≥
2, the sequence S = s , s , ..., s l − of length 3 l − type up-down-up ( l ) if the first l terms are pairwise different and s i = s l − i = s (2 l − i for every i, ≤ i ≤ i . For example, a, b, c, d, c, b, a, b, c, d, is an up-down-up (4) sequence. Valtr and Klazar [14] showed that any l -regular sequence over an n -element alphabet, which contains no subsequence of type up-down-up( l ), has length at most 2 l c n for some constant c . This has been improved by Pettie [20], who proved the following. Lemma 2.2 (Pettie) . For l ≥ , the length of any l -regular sequence over an n -element alphabet,which contains no subsequence of type up-down-up ( l ) , has length at most O ( l ) n . For more results on generalized Davenport-Schinzel sequences, see [15, 20, 19].
In this section, we prove a useful lemma on intersection graphs of curves. It shows that everycollection C of curves, no two of which intersect many times, contains a large subcollection C ′ suchthat in the partition of C ′ into its connected components C , . . . , C t in the intersection graph of C ,each component C i has a vertex connected to all other | C i | − G = ( V, E ), a subset V of the vertex set is said to be a separator if there is apartition V = V ∪ V ∪ V with | V | , | V | ≤ | V | such that no edge connects a vertex in V to avertex in V . We need the following separator lemma for intersection graphs of curves, establishedin [9]. Lemma 3.1 (Fox–Pach) . There is an absolute constant c such that every collection C of curveswith x intersection points has a separator of size at most c √ x . C of curves in the plane decomposable if there is a partition C = C ∪ . . . ∪ C t such that each C i contains a curve which intersects all other curves in C i , and for i = j , the curvesin C i are disjoint from the curves in C j . The following lemma is a strengthening of Proposition6.3 in [8]. Its proof is essentially the same as that of the original statement. It is include here, forcompleteness. Lemma 3.2.
There is an absolute constant c > such that every collection C of m ≥ curvessuch that each pair of them intersect in at most t points has a decomposable subcollection of size atleast cmt log m . Proof of Lemma 3.2
We prove the following stronger statement. There is an absolute constant c > C of m ≥ x edges,and each pair of curves intersect in at most t points, has a decomposable subcollection of size atleast cmt log m + xm . Let c = c , where c ≥ m , noting that all collections of curves with at most three elements are decomposable.Define d = d ( m, x, t ) := cmt log m + xm .Let ∆ denote the maximum degree of the intersection graph of C . We have ∆ < d − C that intersect it, is decomposable and its size is at least d , and we are done. Also, ∆ ≥ xm , since2 xm is the average degree of the vertices in the intersection graph of C . Hence, if ∆ ≥ cmt log m , thenthe desired inequality holds. Thus, we may assume ∆ < cmt log m .Applying Lemma 3.1 to the intersection graph of C , we obtain that there is a separator V ⊂ C with | V | ≤ c √ tx , where c is the absolute constant in Lemma 3.1. That is, there is a partition C = V ∪ V ∪ V with | V | , | V | ≤ | V | / V intersects any curve in V . For i = 1 ,
2, let m i = | V i | and x i denote the number of pairs of curves in V i that intersect, so that x + x ≥ x − ∆ | V | ≥ x − cmt log m c √ tx. (1)As no curve in V intersects any curve in V , the union of a decomposable subcollection of V and adecomposable subcollection of V is decomposable. Thus, by the induction hypothesis, C containsdecomposable subcollection of size at least d ( m , x , t ) + d ( m , x , t ) = cm t log m + x m + cm t log m + x m ≥ c ( m + m ) t log(2 m/
3) + ( x + x )2 m/ . We split the rest of the proof into two cases.
Case 1. x ≥ t − (cid:16) c c m log m (cid:17) . In this case, by (1), we have x + x ≥ x and hence there is adecomposable subcollection of size at least d ( m , x , t ) + d ( m , x , t ) ≥ c ( m + m ) t log m + 5 x m = d + x m − c ( m − ( m + m )) t log m ≥ d + x m − c c √ txt log m > d, completing the analysis. 4 ase 2. x < t − (cid:16) c c m log m (cid:17) . There is a decomposable subcollection of size at least d ( m , x , t ) + d ( m , x , t ) ≥ c ( m + m ) t log(2 m/ ≥ ct (cid:16) m − c √ tx (cid:17) (cid:18) m + 12 log m (cid:19) ≥ ct (cid:18) m log m + m m − c √ tx log m (cid:19) ≥ ct (cid:18) m log m + m m (cid:19) ≥ ct (cid:18) m log m + m m (cid:19) ≥ cmt log m + xm = d, where we used c = c . (cid:3) In this section, we prove Theorem 1.1. The following statement will be crucial for our purposes.
Theorem 4.1.
Let G = ( V, E ) be a k -quasi-planar simple topological graph with n vertices. Supposethat G has an edge that crosses every other edge. Then we have | E | ≤ n · α ( n ) c ′ k , where α ( n ) denotesthe inverse Ackermann function and c ′ k is a constant that depends only on k . Proof of Theorem 4.1.
Let k ≥ c ′ k = 40 · k +2 k . To simplify the presentation, wedo not make any attempt to optimize the value of c ′ k . Label the vertices of G from 1 to n , i.e.,let V = { , , . . . , n } . Let e = uv be the edge that crosses every other edge in G . Note that d ( u ) = d ( v ) = 1.Let E ′ denote the set of edges that cross e . Suppose without loss of generality that no twoof elements of E ′ cross e at the same point. Let e , e , ..., e | E ′ | denote the edges in E ′ listed inthe order of their intersection points with e from u to v . We create two sequences of vertices S = p , p , ..., p | E ′ | and S = q , q , ..., q | E ′ | ⊂ V , as follows. For each e i ∈ E ′ , as we move alongedge e from u to v and arrive at the intersection point with e i , we turn left and move along edge e i until we reach its endpoint u i . Then we set p i = u i . Likewise, as we move along edge e from u to v and arrive at edge e i , we turn right and move along edge e i until we reach its other endpoint w i .Then we set q i = w i . Thus, S and S are sequences of length | E ′ | over the alphabet { , , ..., n } .See Figure 1 for a small example.We need two lemmas. The first one is due to Valtr [23]. Lemma 4.2 (Valtr) . For l ≥ , at least one of the sequences S , S defined above contains an l -regular subsequence of length at least | E ′ | / (4 l ) . (cid:3) Since each edge in E ′ crosses e exactly once, the proof of Lemma 4.2 can be copied almost verbatim from the proof of Lemma 4 in [23] and is left to the reader as an exercise.For the rest of this section, we set l = 2 k + k and t = 2 k . Lemma 4.3.
Neither of the sequences S and S has a subsequence of type up ( l, t ) . Proof.
By symmetry, it suffices to show that S does not contain a subsequence of type up ( l, t ).The argument is by contradiction. We will prove by induction on k that the existence of such asequence would imply that G has k pairwise crossing edges. The base cases k = 1 , k −
1. Let5 (cid:0)(cid:1)(cid:1) (cid:0)(cid:0)(cid:1)(cid:1)(cid:0)(cid:0)(cid:1)(cid:1) (cid:0)(cid:1) (cid:0)(cid:0)(cid:1)(cid:1)(cid:0)(cid:0)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) v v v v v vu Figure 1: In this example, S = v , v , v , v , v and S = v , v , v , v , v . S = s , s , ..., s lt be our up ( l, t ) sequence of length lt such that the first l terms are pairwise distinct and for i =1 , , ..., l we have s i = s i + l = s i +2 l = s i +3 l = · · · = s i +( t − l . For each i = 1 , , ..., l , let v i ∈ V denote the vertex s i . Moreover, let a i,j be the arc emanating fromvertex v i to the edge e corresponding to s i + jl for j = 0 , , , ..., t −
1. We will think of s i + jl as apoint on a i,j very close but not on edge e . For simplicity, we will let s lt + q = s q for all q ∈ N and a i,j = a i,j ′ for all j ∈ Z , where j ′ ∈ { , , , . . . , t − } is such that j ≡ j ′ (mod t ). Hence there are l distinct vertices v , ..., v l , each vertex of which has t arcs emanating from it to the edge e .Consider the arrangement formed by the t arcs emanating from v and the edge e . Since G issimple, these arcs partition the plane into t regions. By the pigeonhole principle, there is a subset V ′ ⊂ { v , ..., v l } of size l − t = 2 k + k − k such that all of the vertices of V ′ lie in the same region. Let j ∈ { , , , ..., t − } be an integersuch that V ′ lies in the region bounded by a ,j , a ,j +1 , and e . See Figure 2. In the case j = t − V ′ lies in the unbounded region.Let v i ∈ V ′ and a i,j + j be an arc emanating from v i for j ≥
1. Notice that a i,j + j cannotcross both a ,j and a ,j +1 , since G is a simple topological graph. Suppose that a i,j + j crosses a ,j +1 . Then all arcs emanating from v i , A = { a i,j +1 , a i,j +2 , ..., a i,j + j − } must also cross a ,j +1 . Indeed, let γ be the simple closed curve created by the arrangement a i,j + j ∪ a ,j +1 ∪ e. Since a i,j + j , a ,j +1 , and e pairwise intersect at precisely one point, γ is well defined. We definepoints x = a i,j + j ∩ a ,j +1 and y = a ,j +1 ∩ e , and orient γ in the direction from x to y along γ .6 (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: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:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(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:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) v s
11, j a vu s s
1, j +1 a V ’ Figure 2: Vertices of V ′ lie in the region enclosed by a ,j , a ,j +1 , e .In view of the fact that a i,j + j intersects a ,j +1 , the vertex v i must lie to the right of γ .Moreover, since the arc from x to y along a ,j +1 is a subset of γ , the points corresponding to thesubsequence S ′ = { s q ∈ S | j + 1) l ≤ q ≤ ( i −
1) + ( j + j ) l } must lie to the left of γ . Hence, γ separates vertex v i and the points of S ′ . Therefore, using againthat G is simple, each arc from A must cross a ,j +1 (these arcs cannot cross a i,j + j ). See Figure4. By the same argument, if the arc a i,j − j crosses a ,j for j ≥
1, then the arcs emanating from v i , a i,j − , a i,j − , ..., a i,j − j +1 must also cross a ,j . Since a i,j + t/ = a i,j − t/ , we have the following observation. Observation 4.4.
For half of the vertices v i ∈ V ′ , the arcs emanating from v i satisfy1. a i,j +1 , a i,j +2 , ..., a i,j + t/ all cross a ,j +1 , or2. a i,j − , a i,j − , ..., a i,j − t/ all cross a ,j . Since t/ k − and | V ′ | ≥ l − t = 2 k + k − · k ≥ ( k − +( k − , by Observation 4.4, we obtain a up (2 ( k − +( k − , k − ) sequence such that the corresponding arcsall cross either a ,j or a ,j +1 . By the induction hypothesis, it follows that there exist k pairwisecrossing edges. (cid:3) Now we are ready to complete the proof of Lemma 4.1. By Lemma 4.2 we know that, say, S contains an l -regular subsequence of length | E ′ | / (4 l ). By Theorem 2.1 and Lemma 4.3, thissubsequence has length at most n · l ( lt − · (10 l ) α ( n ) lt . Therefore, we have 7 (cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:1)(cid:1) v a vu s sa i, j +j a
1, j +1 v S ’ i x y1+j l 1+(j +1) l (a) The case when j + j mod t ≤ t − (cid:0)(cid:0)(cid:1)(cid:1) (cid:0)(cid:1) v a vu s sa i, j +j γ a
1, j +1 v S ’ i x y1+j l 1+(j +1) l (b) γ defined from Figure 3(a). (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) v a i, j +j a S ’ S ’ vu
1, j +1 a v i x y s s (c) The case when j + j mod t < j . Recall a i,j + j = a i,j + j mod 2 k . (cid:0)(cid:1)(cid:0)(cid:0)(cid:1)(cid:1) (cid:0)(cid:1) (cid:0)(cid:0)(cid:1)(cid:1) v a i, j +j a S ’ S ’ vu
1, j +1 a v i γ x y s s (d) γ defined from Figure 3(c). Figure 3: Defining γ and its orientation. | E ′ | · l ≤ n · l ( lt − · (10 l ) α ( n ) lt , which implies | E ′ | ≤ · n · l ( lt − · (10 l ) α ( n ) lt . Since c ′ k = 40 · lt = 40 · k +2 k , α ( n ) ≥ k ≥
5, we have | E | = | E ′ | + 1 ≤ n · α c ′ k ( n ) , which completes the proof of Lemma 4.1. (cid:3) Now we are in a position to prove Theorem 1.1.
Proof of Theorem 1.1.
Let G = ( V, E ) be a k -quasi-planar simple topological graph on n vertices. By Lemma 3.2, there is a subset E ′ ⊂ E such that | E ′ | ≥ c | E | / log | E | , where c is anabsolute constant and E ′ is decomposable. Hence, there is a partition8 ′ = E ∪ E ∪ · · · ∪ E t such that each E i has an edge e i that intersects every other edge in E i , and for i = j , the edges in E i are disjoint from the edges in E j . Let V i denote the set of vertices that are the endpoints of theedges in E i , and let n i = | V i | . By Lemma 4.1, we have | E i | ≤ n i α c ′ k ( n i ) + 2 n i , where the 2 n i term accounts for the edges that share a vertex with e i . Hence, c | E | log | E | ≤ t X i =1 n i α c ′ k ( n i ) + 2 n i ≤ n α c ′ k ( n ) + 2 n, Therefore, we obtain | E | ≤ ( n log n )2 α ck ( n ) , for a sufficiently large constant c k . (cid:3) x -Monotone Topological Graphs The aim of this section is to prove Theorem 1.2.
Proof of Theorem 1.2.
For k ≥
2, let g k ( n ) be the maximum number of edges in a k -quasi-planartopological graph whose edges are drawn as x -monotone curves. We will prove by induction on n that g k ( n ) ≤ ck n log n where c is a sufficiently large absolute constant.The base case is trivial. For the inductive step, let G = ( V, E ) be a k -quasi-planar topologicalgraph whose edges are drawn as x -monotone curves, and let the vertices be labeled 1 , , ..., n . Let L be a vertical line that partitions the vertices into two parts, V and V , such that | V | = ⌊ n/ ⌋ vertices lie to the left of L , and | V | = ⌈ n/ ⌉ vertices lie to the right of L . Furthermore, let E denote the set of edges induced by V , let E denote the set of edges induced by V , and let E ′ bethe set of edges that intersect L . Clearly, we have | E | ≤ g k ( ⌊ n/ ⌋ ) and | E | ≤ g k ( ⌈ n/ ⌉ ) . It suffices that show that | E ′ | ≤ ck / n, (2)since this would imply g k ( n ) ≤ g k ( ⌊ n/ ⌋ ) + g k ( ⌈ n/ ⌉ ) + 2 ck / n ≤ ck n log n. In the rest of the proof, we only consider the edges belonging to E ′ . For each vertex v i ∈ V ,consider the graph G i whose vertices are the edges with v i as a left endpoint, and two vertices9n G i are adjacent if the corresponding edges cross at some point to the left of L . Since G i isan incomparability graph (see [7], [11]) and does not contain a clique of size k , G i contains anindependent set of size | E ( G i ) | / ( k − v i . After repeating this process for allvertices in V , we are left with at least | E ′ | / ( k −
1) edges.Now we continue this process on the other side. For each vertex v j ∈ V , consider the graph G j whose vertices are the edges with v j as a right endpoint, and two vertices in G j are adjacent if thecorresponding edges cross at some point to the right of L . Since G j is an incomparability graphand does not contain a clique of size k , G j contains an independent set of size | E ( G j ) | / ( k − v j .After repeating this process for all vertices in V , we are left with at least | E ′ | / ( k − edges.We order the remaining edges e , e , ..., e m in the order in which they intersect L from bottomto top. (We assume without loss of generality that any two intersection points are distinct.) Definetwo sequences, S = p , p , ..., p m and S = q , q , ..., q m , such that p i denotes the left endpoint ofedge e i and q i denotes the right endpoint of e i . We need the following lemma. Lemma 5.1.
Neither of the sequences S and S has subsequence of type up-down-up ( k + 2) . Proof.
By symmetry, it suffices to show that S does not have a subsequence of type up-down-up ( k + 2). Suppose for contradiction that S does contain such a subsequence. Then there is asequence S = s , s , ..., s k +2) − such that the integers s , ..., s k +2 are pairwise distinct and s i = s k +2) − i = s k +2) − i for i = 1 , , ..., k + 2.For each i ∈ { , , ..., k + 2 } , let v i ∈ V denote the label (vertex) of s i and let x i denote the x -coordinate of the vertex v i . Moreover, let a i be the arc emanating from vertex v i to the point on L that corresponds to s k +2) − i . Let A = { a , a , ..., a k +1 } . Note that the arcs in A are enumerateddownwards with respect to their intersection points with L , and they correspond to the elementsof the “middle” section of the up-down-up sequence. We define two partial orders on A as follows. a i ≺ a j if i < j, x i < x j and the arcs a i , a j do not intersect, a i ≺ a j if i < j, x i > x j and the arcs a i , a j do not intersect.Clearly, ≺ and ≺ are partial orders. If two arcs are not comparable by either ≺ or ≺ , thenthey must cross. Since G does not contain k pairwise crossing edges, by Dilworth’s theorem, thereexist k arcs { a i , a i , ..., a i k } such that they are pairwise comparable by either ≺ or ≺ . Now theproof falls into two cases. Case 1.
Suppose that a i ≺ a i ≺ · · · ≺ a i k . Then the arcs emanating from v i , v i , ..., v i k tothe points corresponding to s k +2) − i , s k +2) − i , ..., s k +2) − i k are pairwise crossing. SeeFigure 4. Case 2.
Suppose that a i ≺ a i ≺ · · · ≺ a i k . Then the arcs emanating from v i , v i , ..., v i k to thepoints corresponding to s i , s i , ..., s i k are pairwise crossing. See Figure 5.10 (cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) i v i v i v i k v i a i a i k a i a (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) i v i v i v i k v i a i a i a i k a sss s k321 Figure 4: Case 1. (cid:3)
We are now ready to complete the proof of Theorem 1.2. By Lemma 4.2, we know that, S ,say, contains a ( k + 2)-regular subsequence of length | E ′ | k + 2)( k − . By Lemmas 2.2 and 5.1, this subsequence has length at most 2 c ′ k n , where c ′ is an absolute constant.Hence, we have | E ′ | k + 2)( k − ≤ c ′ k n, which implies that | E ′ | ≤ k c ′ k n ≤ ck / n for a sufficiently large absolute constant c . (cid:3) (cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) i k v i v i v i v i a i k a i a i a (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) i k v i v i v i v i k s i s i s i s i a i k a i a i a Figure 5: Case 2.
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