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Theorem 1.1. Let G = (V, E) be an n-vertex simple k-quasi-planar graph. Then E(G) 2 αck (n) | | ≤ (n log n)2 , where α(n) denotes the inverse Ackermann function and ck is a constant that depends only on k.
In the proof of Theorem 1.1, we apply results on generalized Davenport-Schinzel sequences. This method was used by Valtr [21], who showed that every n-vertex simple k-quasi-planar graph with ck edges drawn as x-monotone curves has at most 22 n log n edges, where c is an absolute constant. Our next theorem extends his result to (not simple) topological graphs with edges drawn with x-monotone curves, and moreover we obtain a slightly better upper bound.
Theorem 1.2. Let G = (V, E) be an n-vertex (not simple) k-quasi planar graph with edges drawn 3 as x-monotone curves. Then E(G) 2ck n log n, where c is an absolute constant. | |≤ 2 Generalized Davenport-Schinzel sequences
The sequence u = a1, a2, ..., am is called l-regular if any l consecutive terms are pairwise different. For integers l,t 2, the sequence ≥
S = s1,s2, ..., slt of length l t is said to be of type up(l,t) if the first l terms are pairwise different and for i = 1, 2, ..., l · s = s = s = = s . i i+l i+2l · · · i+(t−1)l For example,
a, b, c, a, b, c, a, b, c, a, b, c, would be an up(3, 4) sequence. By applying a theorem of Klazar on generalized Davenport-Schinzel sequences, we have the following.
Theorem 2.1 ([11]). For l 2 and t 3, the length of any l-regular sequence over an n-element ≥ ≥ alphabet that does not contain a subsequence of type up(l,t) has length at most
lt n l2(lt−3) (10l)10α (n). · · For l 2, the sequence ≥
S = s1,s2, ..., s3l−2
2 of length 3l 2 is said to be of type up-down-up(l), if the first l terms are pairwise different, and − for i = 1, 2, ..., l,
si = s2l−i = s(2l−2)+i. For example,
a, b, c, d, c, b, a, b, c, d, would be an up-down-up(4) sequence. Valtr and Klazar showed the following.
Lemma 2.2 ([12]). For l 2, the length of any l-regular sequence over an n-element alphabet ≥ containing no subsequence of type up-down-up(l) has length at most 2O(l)n.
For more results on generalized Davenport-Schinzel sequences, see [13, 18, 17].
3 Simple topological graphs
In this section, we will prove Theorem 1.1. For any partition of V (G) into two disjoint parts, V1 and V2, let E(V1,V2) denote the set of edges with one endpoint in V1 and the other endpoint in V2. The bisection width of a graph G, denoted by b(G), is the smallest nonnegative integer such that there is a partition of the vertex set V = V ˙ V with 1 V V 2 V for i = 1, 2, and 1 ∪ 2 3 · | | ≤ | i|≤ 3 · | | E(V ,V ) = b(G). We will use the following result by Pach et al. | 1 2 | Lemma 3.1 ([16]). If G is a graph with n vertices of degrees d1, ..., dn, then
n b(G) 7cr(G)1/2 + 2v d2, ≤ u i uXi=1 t where cr(G) denotes the crossing number of G.
Since n d2 2n E(G) holds for every graph, we have i=1 i ≤ | | P b(G) 7cr(G)1/2 + 3 E(G) n. (1) ≤ p| | Proof of Theorem 1.1. Let k 5 and f (n) denote the maximum number of edges in a simple ≥ k k-quasi-planar graph on n vertices. We will prove that
c f (n) (n log2 n)2α k (n) k ≤ 2 where c = 105 2k +2k. For sake of clarity, we do not make any attempts to optimize the value of k · c . We proceed by induction on n. The base case n< 7 is trivial. For the inductive step n 7, let k ≥ G = (V, E) be a simple k-quasi-planar graph with n vertices and m = fk(n) edges, such that the vertices of G are labeled 1 to n. The proof splits into two cases. Case 1. Suppose that cr(G) m2/(104 log2 n). By (1), there is a partition V (G) = V V with ≤ 1 ∪ 2 V , V 2n/3 and the number of edges with one vertex in V and one vertex in V is at most | 1| | 2|≤ 1 2
3 m b(G) 7cr(G)1/2 + 3√mn 7 + 3√mn. ≤ ≤ 100 log n Let n = V and n = V . Now if 7m/(100 log n) 3√mn, then we have 1 | 1| 2 | 2| ≤ m 43n log2 n ≤ and we are done since α(n) 2 and k 5. Therefore, we can assume 7m/(100 log n) > 3√mn, ≥ ≥ which implies m b(G) . (2) ≤ 7 log n By the induction hypothesis and equation (2), we have
m f (n )+ f (n )+ b(G) ≤ k 1 k 2 c c n log2(2n/3) 2α k (n) + n log2(2n/3) 2α k (n) + b(G) ≤ 1 2 c n log2(2n/3) 2α k (n) + m ≤ 7 log n c c c (n log2 n)2α k (n) 2n2α k (n) log n log(3/2) + n2α k (n) log2(3/2) + m ≤ − 7 log n which implies
2 1 c 2 log(3/2) log (3/2) m 1 (n log2 n)2α k (n) 1 + . − 7 log n ≤ − log n log2 n Hence
−1 2 −2 c 1 2 log(3/2) log n + log (3/2) log n c m (n log2 n)2α k (n) − (n log2 n)2α k (n). ≤ 1 1/(7 log n) ≤ − Case 2. Now suppose that cr(G) m2/(104 log2 n). By a simple averaging argument, there exists ≥ an edge e = uv such that at least 2m/(104 log2 n) other edges cross e. Fix such an edge e = uv, and let E′ denote the set of edges that cross e. We order the edges in E′ = e , e , ..., e ′ , in the order that they cross e from u to v. Now { 1 2 |E |} we create two sequences S = a , a , ..., a ′ and S = b , b , ..., b ′ as follows. For each e E′, 1 1 2 |E | 2 1 2 |E | i ∈ as we move along edge e from u to v and arrive at edge ei, we turn left and move along edge ei until we reach its endpoint ui. Then we set ai = ui. Likewise, as we move along edge e from u to v and arrive at edge ei, we turn right and move along edge ei until we reach its other endpoint vi. Then set b = v . Thus S and S are sequences of length E′ over the alphabet 1, 2, ..., n . See i i 1 2 | | { } Figure 1 for a small example. Now we need the following two lemmas. The first one is due to Valtr.
Lemma 3.2 ([21]). For l 1, at least one of the sequences S ,S defined above contains an ≥ 1 2 l-regular subsequence of length at least E′ /(4l). | |
4 v
v3 v 4
v v 1 5
v u 2
Figure 1: In this example, S1 = v1, v3, v4, v3, v2 and S2 = v2, v2, v1, v5, v5.
k2+k k Lemma 3.3. Neither of the sequences S1 nor S2 contains a subsequence of type up(2 , 2 ).
k2+k k Proof. By symmetry, it suffices to show that S1 does not contain a subsequence of type up(2 , 2 ). We will prove by induction on k, that such a sequence will produce k pairwise crossing edges in G. The base cases k = 1, 2 are trivial. Now assume the statement holds up to k 1. Let −
2 S = s1,s2, ..., s2k +2k 2 2 2 be our up(2k +k, 2k) sequence of length 2k +2k such that the first 2k +k terms are pairwise different, 2 and for i = 1, 2, ..., 2k +k
s = s 2+ = s 2+ = s 2+ = = s 2+ . i i+2k k i+2·2k k i+3·2k k · · · i+(2k−1)2k k k2 +k For each i = 1, 2, ..., 2 , let vi V1 denote the label (vertex) of si. Moreover, let ai,j be the arc ∈ k emanating from vertex v to the edge e corresponding to s 2 for j = 0, 1, 2, ..., 2 1. We i i+j2k +k − 2 will think of si+j2k +k as a point on ai,j very close but not on edge e. For simplicity, we will let 2 2 N Z k +k s2k +2k+t = st for all t and ai,j = ai,j mod 2k for all j . Hence there are 2 distinct ∈ k ∈ 2 vertices v1, ..., v2k +k , each vertex of which has 2 arcs emanating from it to the edge e. k Consider the drawing of the 2 arcs emanating from v1 and the edge e. This drawing partitions k ′ the plane into 2 regions. By the Pigeonhole principle, there is a subset V v , ..., v 2+ of size ⊂ { 1 2k k } 2 2k +k 1 − , 2k such that all of the vertices of V ′ lie in the same region. Let j 0, 1, 2, ..., 2k 1 be an integer 0 ∈ { − } such that V ′ lies in the region bounded by a , a , e. See Figure 2. In the case j = 2k 1, 1,j0 1,j0+1 0 − V ′ lies in the unbounded region. Let v V ′ and a be an arc emanating out of v for j 1. Notice that a cannot i ∈ i,j0+j1 i 1 ≥ i,j0+j1 cross both a1,j0 and a1,j0+1 since G is simple. Suppose that ai,j0+j1 crosses a1,j0+1. Then the set of arcs (emanating out of vi)
5 v
a1, j +1 v 0 1 ' V a 1, j0 s k 1+(j 0+1) 2 s1+j 2k 0
s1 u
′ Figure 2: Vertices of V lie in the region enclosed by a1,j0 , a1,j0+1, e.
A = a , a , ..., a { i,j0+1 i,j0+2 i,j0+j1−1} must also cross a1,j0+1. Indeed, let γ be the simple closed curve created by the arrangement
a a e. i,j0+j1 ∪ 1,j0+1 ∪
Since ai,j0+j1 , a1,j0+1, e pairwise intersect at precisely one point, γ is well defined. We define points x = a a and y = a e, and orient γ in the direction from x to y along γ. i,j0+j1 ∩ 1,j0+1 1,j0+1 ∩ Since ai,j0+j1 intersects a1,j0+1, vi must lie to the right of γ. Moreover since the arc from x to y along a1,j0+1 is a subset of γ, the points corresponding to the subsequence
2 2 S′ = s S 2 + (j + 1)2k +k t (i 1) + (j + j )2k +k { t ∈ | 0 ≤ ≤ − 0 1 } ′ lie to the left of γ. Hence γ separates vertex vi and the points of S . Since each arc from A must cross γ, each arc must cross a1,j0+1 since G is simple (these arcs cannot cross ai,j0+j1 ). See Figure 3. By the same argument, if the arc a crosses a for j 1, then the arcs (emanating out i,j0−j1 1,j0 1 ≥ of vi)
ai,j0−1, ai,j0−2, ..., ai,j0−j1+1 must also cross a1,j0 . Therefore, we have the following observation. Observation 3.4. For half of the vertices v V ′, the arcs emanating out of v satisfy i ∈ i
1. ai,j0+1, ai,j0+2, ..., ai,j0+2k/2 all cross a1,j0+1, or
2. ai,j0−1, ai,j0−2, ..., ai,j0−2k/2 all cross a1,j0 . Since
′ k2+k V 2 1 2 | | − 2(k−1) +(k−1), 2 ≥ 2 2k ≥ ·
6 a a γ i, j 0+j1 i, j 0+j1
v v
' ' a1, j +1 S a1, j +1 S v 0 v 0 1 1 vi vi a a 1, j0 1, j0 2 2 s k +k s k +k 2 1+(j +1) 2 2 1+(j +1) 2 s k+k 0 s k+k 0 1+j 0 2 1+j 0 2
u u k k (a) The case when j0 + j1 mod 2 ≤ 2 − 1. (b) γ defined from Figure 3(a). a i, j +j1 a 0 i, j 0+j1
v v ' a1, j +1 S a S' v 0 1, j +1 1 v 0 1 vi vi a1, j a 0 1, j0 2 γ k +k k2+k 2 s1+(j +1) 2 s k 0 2 1+(j 0+1) 2 s1+j 2 +k s1+j 2 k +k 0 0 ' S S'
u u k (c) The case when j0 + j1 mod 2 < j0. Recall (d) γ defined from Figure 3(c). k ai,j0+j1 = ai,j0+j1 mod 2 .
Figure 3: Defining γ and its orientation.
2 by Observation 3.4 we have an (2(k−1) +(k−1), 2k−1)up sequence, whose corresponding arcs all cross either a1,j0 or a1,j0+1. By the induction hypothesis, we have k pairwise crossing edges.
Now we are ready to complete the proof of Theorem 1.1. By Lemma 3.2 we know that, say, S1 2 2 contains a 2k +k-regular subsequence of length E′ /(4 2k +k). By Theorem 2.1 and Lemma 3.3, | | · this subsequence has length at most
2 2k +2k 2 k2+2k 2 10α (n) n2k +k22 −3 10 2k +k . · Therefore
2 ′ 2k +2k 2m E 2 k2+2k 2 10α (n) | | n2k +k22 −3 10 2k +k 104 4 2k2+k log2 n ≤ 4 2k2+k ≤ · · · · which implies
2 2k +2k 2 k2+2k 2 10α (n) m 4 104 22k +2k22 −3n 10 2k +k log2 n. ≤ · · · 7 2 Since c = 105 2k +2k, α(n) 2 and k 5, we have k · ≥ ≥ c m (n log2 n)2α k (n). ≤
4 x-monotone
In this section we will prove Theorem 1.2. Proof of Theorem 1.2. For k 2, let g (n) be the maximum number of edges in a (not simple) ≥ k k-quasi-planar graph whose edges are drawn as x-monotone curves. We will prove by induction on n that
3 g (n) 2ck n log n k ≤ where c is a sufficiently large absolute constant. The base case is trivial. For the inductive step, let G = (V, E) bea k-quasi-planar topological graph whose edges are drawn as x-monotone curves, and let the vertices be labeled 1, 2, ..., n. Then let L be the vertical line that partitions the vertices into two parts, V and V , such that V = n/2 vertices lie to the left of L, and V = n/2 1 2 | 1| ⌊ ⌋ | 2| ⌈ ⌉ vertices lie to the right of L. Furthermore, let E1 denote the set of edges induced by V1, E2 be the ′ set of edges induced by V2, and E be the set of edges that intersect L. Clearly, we have
E g ( n/2 ) and E g ( n/2 ). | 1|≤ k ⌊ ⌋ | 2|≤ k ⌈ ⌉ Hence it suffices that show that
3 E′ 2ck /2n, (3) | |≤ since this would imply
3 3 g (n) g ( n/2 )+ g ( n/2 ) + 2ck /2n 2ck n log n. k ≤ k ⌊ ⌋ k ⌈ ⌉ ≤ For the rest of the proof, we will only consider the edges from E′. Now for each vertex v V , i ∈ 1 consider the graph Gi whose vertices are the edges with vi as a left endpoint, and two vertices in Gi are adjacent if the corresponding edges cross at some point to the left of L. Since Gi is an incomparability graph (see [7], [9]) and does not contain a clique of size k, Gi contains an independent set of size E(G ) /(k 1). We keep all edges that correspond to the elements of this | i | − independent set, and discard all other edges incident to vi. After repeating this process on all vertices in V , we are left with at least E′ /(k 1) edges. 1 | | − Now we continue this process on the other side. For each vertex v V , consider the graph G j ∈ 2 j whose vertices are the edges with vj as a right endpoint, and two vertices in Gj are adjacent if the corresponding edges cross at some point to the right of L. Since Gj is an incomparability graph and does not contain a clique of size k, G contains an independent set of size E(G ) /(k 1). We j | j | − keep all edges that corresponds to this independent set, and discard all other edges incident to vj. After repeating this process on all vertices in V , we are left with at least E′ /(k 1)2 edges. 2 | | − We order the remaining edges e1, e2, ..., em in the order in which they intersect L from bottom to top. We define two sequences S1 = a1, a2, ..., am and S2 = b1, b2, ..., bm such that ai denotes the left endpoint of edge ei and bi denotes the right endpoint of ei. Now we need the following lemma.
8 3 Lemma 4.1. Neither of the sequences S1 or S2 contains a subsequence of type up-down-up(k +2).
Proof. By symmetry, it suffices to show that S1 does not contain a subsequence of type up- 3 down-up(k + 2). For the sake of contradiction, suppose S1 did contain a subsequence of type up-down-up(k3 + 2). Then there is a sequence
S = s1,s2, ..., s3(k3+2)−2 3 such that the integers s1, ..., sk3+2 are pairwise different and for i = 1, 2, ..., k + 2 we have
si = s2(k3+2)−i = s2(k3+2)−2+i. For each i = 1, 2, ..., k3 + 2, let v V denote the label (vertex) of s and let x denote the i ∈ 1 i i x-coordinate of vertex vi. Moreover, let ai be the arc emanating from vertex vi to the point on L that corresponds to s 3 . Note that the set of arcs A = a , a , ..., a 3 are ordered 2(k +2)−i { 2 3 k +1} downwards along L, and corresponds to the “middle” part of the up-down-up sequence. We define two partial orders on A as follows.
a a if i
s 3 2(k +2)−2+ik s 3 2(k +2)−2+i3 s 3 2(k +2)−2+i2
s 3 2(k +2)−2+i1
ai1 vi ai1 1 vi ai 2 2 v i1 a v i2 i2 a v i i 3 a 3 i3 v vi i k 3 vik ai k aik
Figure 4: Case 1.
9 v i1 ai v v 1 i1 i2 ai1 a vi i2 2 vi 3 a a i v i2 3 i3 a i v 3 i k v i k ai k
aik
si k
s i3 s i2
si1
Figure 5: Case 2.
Case 2. Suppose that a a a . Then the arcs emanating from v , v , ..., v to the i1 ≺2 i2 ≺2 ···≺2 ik i1 i2 ik points corresponding to si1 ,si2 , ..., sik are pairwise crossing. See Figure 5. We are now ready to complete the proof of Theorem 1.2. By Lemma 3.2, we know that, say, 3 S1 contains a (k + 2)-regular subsequence of length
E′ | | . 4(k3 + 2)(k 1)2 − ′ 3 By lemma 2.2 and 4.1, this subsequence has length at most 2c k n, where c′ is an absolute constant. Hence
′ E ′ 3 | | 2c k n 4(k3 + 2)(k 1)2 ≤ − implies
′ 3 3 E′ 4k52c k n 2ck /2n | |≤ ≤ for a sufficiently large absolute constant c.
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