Computational Geometry Seminar

Computational GeometrySeminar Lecture 9 Eyal Zur

These edges are in convex position • Definitions • Two edges of a geometric graph are in convex position if they are disjoint edges of a convex quadrilateral. • A geometric graph is called proper if it has two edges in convex position. • Otherwise, it is called improper. Also … These edges aren’t in convex position

Our motivation Kupitz Conjecture An improper geometric graph with n vertices has at most 2n-2 edges. An improper geometric graph on 7 vertices and 12 edges.

What we are going to see? Theorem (Katchalski and Last) An improper geometric graph with n vertices has at most 2n-1 edges.

Definitions • A circular sequence is a sequence whose first and last term are considered adjacent. • (1,2,3,4,5) is a circular sequence  1 and 5 are adjacent. • A circular sequence from a set of n symbols shall be called a circular Davenport–Schinzel sequence of order 2 if: • No two adjacent terms are identical • and • If it does not contain a circular subsequence of type “abab”. • (a,b,c,d) is a sequence like this, • (1,2,3,4,2,3,1) , (1,2,3,2,3,4) , (a,b,c,a,b) – aren’t

Definition (cont.) • A convex curve is the boundary of a compact convex planar set with nonempty interior. • (v1,v2,v4,v5,v6,v8,v0) – is convex curve • (v1,v2,v3) , (v3,v4,v5,v6) - aren’t v8 v0 v9 v1 v7 v3 v4 v2 v5 v6

We shall need a known upper bound on the length of such circular sequences of order 2 which is proved by induction on n, the number of symbols: Lemma 1 The length of a circular Davenport–Schinzel sequence of order 2 on n symbols for n ≥ 2 is at most 2n-2. not our main issue, so without proof.

Ak Q Aj Ai P Al Lemma 2 Let Ai, Aj, Ak, Al be four points appearing in this order on a convex curve δ. Let P, Q be two points inside δ. Consider the four (closed) segments PAi, QAj, PAk, QAl, and assume that among them: there is no segment s such that s contains only one of the points P, Q and the line supporting s contains both of them. Then two of the four segments are in convex position. (2) Implies that if one of the points P,Q lies on one of the gray lines, then the other one also lies on the same gray line. (1) (2)

B A D C Observation If points A, B, C, D in general position lie in this order on a convex curve, then the segments AB and CD are in convex position. (3)

Proof of lemma 2 Let l=l(P,Q) and let l1and l2 be two half-planes such that l1l2 = l. Let δ1(δ2) be the boundary of the convex hull of l1δ (of l2δ). Each δi contains l also. l Aj Ak Q δ1 l2 l1 δ2 Ai P Al

Proof of lemma 2 (cont.) It is easy to check, using (2) and (3) that if one of the four points Ai, Aj, Ak, Al lies on l, then two of the segments are in convex position. Aj Ak Aj Ak l1 l δ1 l1 l δ1 l2 l2 Q δ2 δ2 Q Ai P Ai P Al Al Here QAj and PAk and are in convex position Here QAl and PAi and are in convex position

Proof of lemma 2 (cont.) • Assume therefore without loss of generality that either: • Ai, Aj and Ak lie in the interior of l1 • And • Ai lie in the interior of l2 • Or • Ai and Aj lie in the interior of l1 • And • Ak and Al lie in the interior of l2 (4) (5)

Proof of lemma 2 (cont.) In case (4), by (3) applied to δ1, either PAi and QAj are in convex position or PAk and QAj are in convex position. l l Aj Aj Ai Ai l1 δ1 δ1 l1 Q l2 l2 P δ2 δ2 P Al Al Q Ak Ak Here QAj and PAi and are in convex position Here QAj and PAk and are in convex position

Proof of lemma 2 (cont.) In case (5) if PAi , QAj are not in convex position and PAk, QAl are not in convex position, then by (3) the order of the points on δ is Ai, Aj, Al, Ak , a contradiction. l Aj Ai l Aj Ai δ1 δ1 l1 l1 Q l2 P l2 δ2 δ2 Al Al Q P Ak Ak

Proof of the Theorem Let v1,…,vn be the vertices of G and e the number of its edges. Assume that G is improper with e ≥ 1. Let C be a circle containing v1,…, vn in its interior. For any two vertices vi and vj joined by an edge vivj define two points on C: and

Proof of the Theorem (cont.) Arrange the 2e points on C in a circular sequence according to the order of their appearance on C. Let D(G) be the circular sequence thus obtained. D(G) = (α41, α42, α43, α23, α12, α14, α24, α34, α32, α21)

Proof of the Theorem (cont.) Color the points of D(G) with n colors such that αij receives the color i : Point αij has a dark color i if vivj is an interior edge of vj. Otherwise, αij has a light color i . D(G) = (α41, α42, α43, α23, α12, α14, α24, α34, α32, α21)

Proof of the Theorem (cont.) Divide the sequence D(G) into arcswhere an arc is a maximal subsequence of consecutive points of D(G) having the same color i . Note that a dark i and a light i may belong to the same arc. For our example: D(G) = (α41,α42,α43,α23,α12,α14,α24,α34,α32,α21)  The arcs of G are (α41, α42, α43),(α23),(α12, α14),(α24),(α34, α32),(α21)

Proof of the Theorem (cont.) The circular sequence obtained from D(G) by contracting each arc to one of its points and then replacing the point by its color i is called the pattern sequence of G, or PS(G). For our example: PS(G) = (4,2,1,2,3,2)

Lemma 3 PS(G) is a circular Davenport-Schinzel sequece of order 2. Lemma 4 An arc of D(G) contains at most one point with dark color. We’ll see proofs for the lemmas soon… But for now we have, |D(G)| = 2e = # of light colored points + # of dark colored points (6)

Proof of the Theorem (cont.) Each vertex of G has at most one rightmost edge and at most one leftmost edge incident to it, so that the number of light colored points in D(G) is bounded by 2n. By Lemmas 1, 3, and 4, # of dark colored points ≤ |PS(G)| ≤ 2n-2 Substitute all of the above in (6) to obtain e ≤ 2n – 1 D(G)| = 2e = # of light colored points + # of dark colored points ≤ 2n + 2n – 2 = 4n –2 It remains to prove lemmas 3 and 4.

αb2 αa1 u2 u1 αa3 αb4 vb u4 u3 va Proof of lemma 3 If PS(G) is not a circular Davenport–Schinzel sequence of order 2, then there are four points, αa1, αb2, αa3, αb4 appearing in that order on C. Since v1,…,vnare in general position there are no two disjoint segments vxvy, vzvtsuch that a line through one of them contains an endpoint of the other. (If exists then there are 3 edges on the same line, and the vertices aren’t in general position)

αb2 αa1 u2 u1 αa3 αb4 vb u4 u3 va Proof of lemma 3 (cont.) Therefore by Lemma 2, two of the segments vaαa1, vbαb2, vaαa3, vbαb4 , are in convex position. Since every edge vxvyis contained in the segment vxαxy, two of the segments vau1, vbu2, vau3, vbu4 Are in convex position, a contradiction. In the example: vau3 and vbu4 are in convex position

Proof of lemma 4 Suppose that the points αab, αacbelong to the same arc and are dark colored. Assume without loss of generality that vavcis to the right of vavb. The edges vavb, vavcare interior edges of vband vc, respectively. So let vbvxbe to the right of vbva and let vcvybe to the left of vcva.

Proof of lemma 4 (cont.) Let stbe the chord of C that contains vbvc, so that vb lies in vcs and vcin vbt. Let r be the ray with apex vband parallel to vcva. Let α1, α2, α3, α4 be the following angles: α1 = conv(vbαbaυ r) α2 = conv(r υ vbs) α3 = conv(vbs υ vbαab) α4 = conv(vct υ vcαac)

Proof of lemma 4 (cont.) Since vbvx is to the right of vbva, vx must lie in at least one of the angles α1 or α2or α3. If vxα1, then the points αac, αxb,αab,αbx are in that order on C. Therefore αab,αac are not on the same arc, a contradiction. If vxα2, then vbvx and vcva are in convex position, a contradiction. Therefore vbvx is in α3 and by symmetry vcvy is in α4, implying vbvx and vcvy are in convex position, a contradiction.

The first end… Get out for a BREAK

Definition • Two edges of a geometric graph are said to be parallelif they are opposite sides of a convex quadrilateral. • Actually (as you see) there is no different • between parallel edges and edges in convex • position. These edges are parallel Also … These edges aren’t parallel

What we are going to see? Theorem (Valtr) Let k ≥ 2 be a constant, Then any geometric graph on n vertices with no k pairwise parallel edges has at most O(n) edges. 9

Generalized Davenport–Schinzel Sequences, definitions: • For l≥ 1, a sequence is called l-regular, if any l consecutive • terms are pairwise different. • Examples: • (1,2,3,4,5) – is 4-regular sequence. • (1,2,3,4,2,5) – isn’t 4-regular sequence, but is 3-regular. • For l ≥ 2, a sequence S = s1, s2, …, s3l-2 • of length 3l-2 is said to be of typeup-down-up(l), • if the first l terms are pairwise different and, • for i=1,2,…,l : si= s2l-i= s(2l-2)+i • Example: (1,2,3,4,5,4,3,2,1,2,3,4,5) – is up-down-up(5) sequence.

Conclusion A sequence is of type up-down-up(2)  it is an alternating sequence of length 4. Meaning, in the form of (a,b,a,b). Claim It is known (and without proof here) that any 2-regular sequence over an n-element alphabet containing no alternating subsequence of length 4 has length at most 2n-1.

In the proof of Theorem 1 we apply the following related result: Theorem 4 Let l≥ 2 be a constant. Then the length of any l-regular sequence over an n-element alphabet containing no subsequence of type up-down-up(l) is at most O(n). (also not our main issue)

Proof of Theorem 1 • Let G=)V,E) be a geometric graph on n vertices, with no k pairwise parallel edges. • Let V=(v1,v2,…,vn). • Without loss of generality, we assume that no two points lie on a horizontal line. If necessary, • we perturb the vertices of G • to make the directions of edges • of G pairwise different. We don’t allow the • red edges exist that way

e1 e3 e4 e2 e5 e6 v6 v8 v1 v5 v4 v7 v2 v3 Proof of theorem 1 (cont.) Let e E. An oriented edge e is defined as the edge e oriented upward. The direction of e, dir(e), is defined as the direction of the vector vivj, where e = (vi,vj), Let E = {e1,e2,…,em), where 0 < dir(e1) < dir(e2) < … < dir(em) <

e1 e3 e4 e2 e5 e6 v6 v8 v1 v5 v4 v7 v2 v3 Proof of theorem 1 (cont.) Definition Let P1 and P2 be the sequences of m integers obtained from the sequence e1, e2, …, em by replacing each edge ek = (vi,vj) by integer i and by integer j , respectively. We call the sequences P1, P2 the pattern sequences of G. P1=(5,3,7,4,2,3) P2=(6,8,6,8,1,2)

Lemma 5 For each l≥ 1, at least one of the pattern sequences P1, P2 contains an l-regular subsequence of length at least |E|/(4l) = m/(4l). Lemma 6 Neither of the pattern sequences P1, P2 contains a subsequence of type up-down-up(k ). Before proving Lemmas 5 and 6, we complete the proof of Theorem 1. 3

Proof of Theorem 1 (cont.) According to Lemma 5, at least one of the sequences P1, P2 contains a k -regular subsequence S of length at least |E|/(4k ). According to Lemma 6, the sequence S contains no subsequence of type up-down-up(k ). Theorem 4 implies that the length of S is at most O(n). Consequently, |E| ≤ 4k O(n) = O(n). It remains to prove Lemmas 5 and 6. 3 3 3 3

Proof of lemma 5 We apply a simple greedy algorithm which, for given integer l ≥ 1 and finite sequence A, returns an l-regular subsequence B(A,l) of A. In the first step, an auxiliary sequence B is taken empty. Then the terms of A are considered one by one from left to right, and in each step the considered term is placed at the end of B iff this does not violate the l-regularity of B. Finally, the obtained l-regular subsequence B of A is taken for B(A,l). For example: A=(1,3,1,3,5,2,2,5,1,5,1,2) and l=3, then the algorithm returns the sequence: B(A,3)=(1,3,5,2,1,5,2).

Proof of lemma 5 (cont.) Let l≥ 1 be given. For i = 1,2 and for 1 ≤ j1≤ j2≤ m, let Pi,[j1,j2] denote the part of Pi starting with the j1th term and ending with the j2th term. Thus, Pi,[ j1, j2 ] consists of j2-j1+1 terms. Let |T| denote the length of a sequence T , and I[T] the set of integers (different numbers) appearing in T . For example: A=(1,3,1,3,5,2,2,5,1,5,1,2) |A| = 12 I[A] = {1,3,5,2}

Claim 7 For each j = 1, 2, …, m : |B(P1,[1, j], l)| + |B(P2,[1, j], l)| ≥ j / (2l) Proof First, consider two integers j1,j2 such that 1≤j1≤j2≤m. Obviously, {ej1,ej1+1 …,ej2} {(va,vb)|a I(P1,[j1,j2]),b I(P2,[j1,j2])}  |{ej1,ej1+1 …,ej2}| ≤ |{(va,vb)|a I(P1,[j1,j2]),b I(P2,[j1,j2])}|  j2-j1+1 ≤ |I(P1,[j1,j2])| x |I(P2,[j1,j2])|

Proof Claim 7 (cont.) By the inequality between algebraic and geometric means, |I(P1,[j1,j2])|+|I(P2,[j1,j2])| (1) ≥ j2-j1+1 2 We can now prove the claim by induction on j. If j ≤ min(16l ,m), then by (1) and by j ≤ 16l |B(P1,[1,j],l)| + |B(P2,[1,j],l)| ≥ |I(P1,[1,j])| + |I(P2,[1,j])| ≥ 2 j = 2j/ j = 2j/4l ≥ j/(2l) 2 2

Proof Claim 7 (cont.) Consequently, by the inductive hypothesis and by (1), |B(P1,[1,j0],l)| + |B(P2,[1,j0],l)| ≥ (j0-4l )/(2l) + 2 4l - 2(l-1) ≥ j0/(2l) 2 2 Proof of Lemma 5 Lemma 5 follows easily from claim 7 (with j = m) and from the pigeon-hole principle: B(P1,[1,m],l)| + |B(P2,[1,m], l)| ≥ m / (2l) , And therefore, at least one of the l-regular sequences is at length of m/(4l).

Proof of Lemma 6 • First, we apply the following consequence of Dilworth’s • Theorem: • Theorem 8 • If the union of three partial orderings on a set I of size at least • (k-1) +1 is a linear ordering on I , then at least one of the partial orderings contains a chain of length k. • Remainder: • By linear ordering we mean: • If a≤b and b≤a then a = b ( antisymmetry) • If a≤b and b≤c then a≤c (transitivity) • a≤b or b≤a (totally). 3

1 1 1 2 2 2 3 3 3 Proof of Theorem 8 Let , , , be the three partial orderings on I. If (I, ) does not contain a chain of length k then, by Dilworth’s theorem, it can be covered by at most k-1 antichains. Consequently, There is an antichain A of size (k-1) +1 in (I, ). If we restrict our attention to A and to orderings , another application of Dilworth’s theorem gives k elements in A which form a chain with respect to or to . (but it’s not our main issue…) 2

Proof of Lemma 6 Because of symmetry, it suffices to prove Lemma 6 for the pattern sequence P1. Suppose to the contrary that P1 contains a subsequence of type up-down-up(k ). Thus, there is a subsequence S = s1, s2, …, s3k -2 of P1 such that the integers s1,s2,…sk are pairwise different and that, for i=1,2,…,k , si=s2k -i=s(2k -2)+i . For simplicity of notation, suppose that si=i (i=1,…,k ) and that S=P1,[1,3k -2]. We obtain a contradiction by showing that k of the edges e1,e2,…,e3k -2 are pairwise parallel. 3 3 3 3 3 3 3 3

1 1 1 2 2 2 3 3 3 • Definition 9 • Let i,j I, and let dir(vi,vj) denote the direction of the vector vi,vj. • Then: • (i) i j, if i<j and dir(vi,vj) [dir(ek ), ] • (ii) i j, if i<j and dir(vi,vj) [ , + dir(e2k -1)] • i j, if i<j and dir(vi,vj) [ + dir(e2k-1), 2 ] U • (0, dir(ek )) 3 3 3 3 Since the union of , , , is a linear ordering on I, Theorem 8 implies that one of the orderings , , contains a chain i1,i2,..,ikof length k. We distinguish the corresponding three possible cases.

1 1 1 2 2 2 3 3 3 Definition 9 (cont.) If i1 i2 …. ik, then the edges ei1, ei2, …, eik are pairwise parallel. Indeed, if 1 ≤ j < j’ ≤ k then the inequalities 0 ≤ dir(eij) < dir(eij’) ≤ dir(ek ) ≤ dir(vij,vij’) < show that the edges eij, eij’ are parallel. 3 Similarly, if i1 i2 … ik, then the edges e(2k -2)+i1,e(2k -2)+i2,…,e(2k -2)+ik are pairwise parallel, and if i1 i2 … ik then the edges e2k –i, e2k -i,…,e2k -i are pairwise parallel. eij’ 3 3 3 vij’ 3 3 3 eij vij

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Computational Geometry Seminar