1. 1 Computational Geometry Seminar Lecture 9 Eyal Zur

2. 2 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. These edges are in convex position Also … These edges aren ’ t in convex position

3. 3 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.

4. 4 What we are going to see? Theorem (Katchalski and Last) An improper geometric graph with n vertices has at most 2n-1 edges.

5. 5 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

6. 6 Definition (cont.) A convex curve is the boundary of a compact convex planar set with nonempty interior. (v 1,v 2,v 4,v 5,v 6, v 8,v 0 ) – is convex curve (v 1,v 2,v 3 ), (v 3,v 4,v 5,v 6 ) - aren’t v8 v7 v0 v6 v2 v5 v1 v4 v3 v9

7. 7 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.

8. 8 Lemma 2 Let A i, A j, A k, A l 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 PA i, QA j, PA k, QA l, 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) AjAj AiAi AkAk AlAl P Q

9. 9 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) A D B C

10. 10 Proof of lemma 2 Let l=l(P,Q) and let l 1 and l 2 be two half-planes such that l 1 l 2 = l. Let δ1(δ2) be the boundary of the convex hull of l 1 δ (of l 2 δ). Each δ i contains l also. AjAj AiAi AkAk AlAl P Q l l1l1 l2l2 δ2δ2 δ1δ1

11. 11 Proof of lemma 2 (cont.) It is easy to check, using (2) and (3) that if one of the four points A i, A j, A k, A l lies on l, then two of the segments are in convex position. AjAj AiAi AkAk AlAl P Q l l1l1 l2l2 δ2δ2 δ1δ1 AjAj AiAi AkAk AlAl P Q l l1l1 l2l2 δ2δ2 δ1δ1 Here QA j and PA k and are in convex position Here QA l and PA i and are in convex position

12. 12 Proof of lemma 2 (cont.) Assume therefore without loss of generality that either: A i, A j and A k lie in the interior of l 1 And A i lie in the interior of l 2 Or A i and A j lie in the interior of l 1 And A k and A l lie in the interior of l 2 (4) (5)

13. 13 Proof of lemma 2 (cont.) In case (4), by (3) applied to δ 1, either PA i and QA j are in convex position or PA k and QA j are in convex position. AiAi AlAl AjAj AkAk Q P l l1l1 l2l2 δ2δ2 δ1δ1 AiAi AlAl AjAj AkAk Q P l l1l1 l2l2 δ2δ2 δ1δ1 Here QA j and PA i and are in convex position Here QA j and PA k and are in convex position

14. 14 Proof of lemma 2 (cont.) In case (5) if PA i, QA j are not in convex position and PA k, QA l are not in convex position, then by (3) the order of the points on δ is A i, A j, A l, A k, a contradiction. AiAi AlAl AjAj AkAk Q P l l1l1 l2l2 δ2δ2 δ1δ1 AiAi AlAl AjAj AkAk P Q l l1l1 l2l2 δ2δ2 δ1δ1

15. 15 Proof of the Theorem Let v 1,…,v n 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 v 1,…, v n in its interior. For any two vertices v i and v j joined by an edge v i v j define two points on C: and

16. 16 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) = (α 41, α 42, α 43, α 23, α 12, α 14, α 24, α 34, α 32, α 21 )

17. 17 Proof of the Theorem (cont.) Color the points of D(G) with n colors such that α ij receives the color i : Point α ij has a dark color i if v i v j is an interior edge of v j. Otherwise, α ij has a light color i. D(G) = ( α 41, α 42, α 43, α 23, α 12, α 14, α 24, α 34, α 32, α 21 )

18. 18 Proof of the Theorem (cont.) Divide the sequence D(G) into arcs where 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) = (α 41,α 42,α 43,α 23,α 12,α 14,α 24,α 34,α 32,α 21 )  The arcs of G are (α 41, α 42, α 43 ),(α 23 ),(α 12, α 14 ),(α 24 ),(α 34, α 32 ),(α 21 )

19. 19 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)

20. 20 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)

21. 21 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.

22. 22 Proof of lemma 3 If PS(G) is not a circular Davenport – Schinzel sequence of order 2, then there are four points, α a1, α b2, α a3, α b4 appearing in that order on C. Since v 1, …,v n are in general position there are no two disjoint segments v x v y, v z v t such 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) α b2 α a3 α a1 α b4 u2u2 u1u1 u3u3 u4u4 vbvb vava

23. 23 Proof of lemma 3 (cont.) Therefore by Lemma 2, two of the segments v a α a1, v b α b2, v a α a3, v b α b4, are in convex position. Since every edge v x v y is contained in the segment v x α xy, two of the segments v a u 1, v b u 2, v a u 3, v b u 4 Are in convex position, a contradiction. In the example: v a u 3 and v b u 4 are in convex position α b2 α a3 α a1 α b4 u2u2 u1u1 u3u3 u4u4 vbvb vava

24. 24 Proof of lemma 4 Suppose that the points α ab, α ac belong to the same arc and are dark colored. Assume without loss of generality that v a v c is to the right of v a v b. The edges v a v b, v a v c are interior edges of v b and v c, respectively. So let v b v x be to the right of v b v a and let v c v y be to the left of v c v a.

25. 25 Proof of lemma 4 (cont.) Let st be the chord of C that contains v b v c, so that v b lies in v c s and v c in v b t. Let r be the ray with apex v b and parallel to v c v a. Let α 1, α 2, α 3, α 4 be the following angles: υ α 1 = conv(v b α ba υ r) υ α 2 = conv(r υ v b s) υ α 3 = conv(v b s υ v b α ab ) υ α 4 = conv(v c t υ v c α ac )

26. 26 Proof of lemma 4 (cont.) Since v b v x is to the right of v b v a, v x must lie in at least one of the angles α 1 or α 2 or α 3. If v x α 1, then the points α ac, α xb, α ab, α bx are in that order on C. Therefore α ab, α ac are not on the same arc, a contradiction. If v x α 2, then v b v x and v c v a are in convex position, a contradiction. Therefore v b v x is in α 3 and by symmetry v c v y is in α 4, implying v b v x and v c v y are in convex position, a contradiction.

27. 27 The first end… Get out for a BREAK

28. 28 Definition Two edges of a geometric graph are said to be parallel if 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

29. 29 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

30. 30 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 = s 1, s 2, …, s 3l-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 : s i = s 2l-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.

31. 31 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.

32. 32 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)

33. 33 Proof of Theorem 1 Let G=)V,E) be a geometric graph on n vertices, with no k pairwise parallel edges. Let V=(v 1,v 2,…,v n ).  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

34. 34 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 v i v j, where e = (v i,v j ), Let E = {e 1,e 2,…,e m ), where 0 < dir(e 1 ) < dir(e 2 ) < … < dir(e m ) < v6v6 v8v8 v1v1 e1e1 e3e3 e4e4 e2e2 e5e5 e6e6 v3v3 v4v4 v2v2 v5v5 v7v7

35. 35 Proof of theorem 1 (cont.) Definition Let P 1 and P 2 be the sequences of m integers obtained from the sequence e 1, e 2, …, e m by replacing each edge e k = (v i,v j ) by integer i and by integer j, respectively. We call the sequences P 1, P 2 the pattern sequences of G. P 1 =(5,3,7,4,2,3) P 2 =(6,8,6,8,1,2) v6v6 v8v8 v1v1 e1e1 e3e3 e4e4 e2e2 e5e5 e6e6 v3v3 v4v4 v2v2 v5v5 v7v7

36. 36 Lemma 5 For each l ≥ 1, at least one of the pattern sequences P 1, P 2 contains an l-regular subsequence of length at least |E|/(4l) = m/(4l). Lemma 6 Neither of the pattern sequences P 1, P 2 contains a subsequence of type up-down-up(k ). Before proving Lemmas 5 and 6, we complete the proof of Theorem 1. 3

37. 37 Proof of Theorem 1 (cont.) According to Lemma 5, at least one of the sequences P 1, P 2 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

38. 38 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).

39. 39 Proof of lemma 5 (cont.) Let l ≥ 1 be given. For i = 1,2 and for 1 ≤ j 1 ≤ j 2 ≤ m, let P i,[j 1,j 2 ] denote the part of P i starting with the j 1 th term and ending with the j 2 th term. Thus, P i,[ j 1, j 2 ] consists of j 2 -j 1 +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}

40. 40 Claim 7 For each j = 1, 2, …, m : |B(P 1,[1, j], l)| + |B(P 2,[1, j], l)| ≥ j / (2l) Proof First, consider two integers j 1,j 2 such that 1≤j 1 ≤j 2 ≤m. Obviously, {e j 1,e j 1+1 …,e j 2 } {(v a,v b )|a I(P 1,[j 1,j 2 ] ),b I(P 2,[j 1,j 2 ] )}  |{e j 1,e j 1+1 …,e j 2 }| ≤ |{(v a,v b )|a I(P 1, [j 1, j 2 ] ),b I(P 2,[j 1,j 2 ] )}|  j 2 -j 1 +1 ≤ |I(P 1,[j 1,j 2 ] )| x |I(P 2,[j 1,j 2 ] )|

41. 41 Proof Claim 7 (cont.) By the inequality between algebraic and geometric means, |I(P 1,[j1,j2] )| + |I(P 2,[j1,j2] )| 2 ≥ j 2 -j 1 +1 (1) We can now prove the claim by induction on j. If j ≤ min(16l,m), then by (1) and by j ≤ 16l |B(P 1,[1,j],l)| + |B(P 2,[1,j],l)| ≥ |I(P 1,[1,j] )| + |I(P 2,[1,j] )| ≥ 2 j = 2j/ j= 2j/4l ≥ j/(2l) 2 2

42. 42 Proof Claim 7 (cont.) Suppose now that 16l < j 0 ≤ m and that Claim 7 holds for j = 1, 2, …, j 0 -1. Since for i=1,2 each integer of I(P i,[j 0 -4l +1, j0] ) not appearing among the last l-1 terms in B(P i,[1,j 0 -4l ],l) appears more times in B(P i,[1, j0],l) than in B(P i,[1, j 0 -4l ],l), we have |B(P i,[1,j0],l)| ≥ |B(P i,[1,j 0 -4l ],l)| + |I(P i,[j 0 -4l +1, j0] )| - (l-1) From our the greedy algorithm, it’s obvious that: |B(P i,[1,j 0 ], l)| ≥ |B(P i,[1,j 0 -4l ],l)| And if |I(P i,[j 0 -4l +1, j 0 ] )| contains more than l-1 integers, then each integer which not appear in |B(P i,[1,j 0 -4l ], l)| adds at least one to the equation ’ s left side. 2 2 2 2 2 2 2 2 2

43. 43 Proof Claim 7 (cont.) Consequently, by the inductive hypothesis and by (1), |B(P 1,[1,j0],l)| + |B(P 2,[1,j0],l)| ≥ (j 0 -4l )/(2l) + 2 4l - 2(l-1) ≥ j 0 /(2l) 2 2 Proof of Lemma 5 Lemma 5 follows easily from claim 7 (with j = m) and from the pigeon-hole principle: B(P 1,[1,m], l)| + |B(P 2,[1,m], l)| ≥ m / (2l), And therefore, at least one of the l-regular sequences is at length of m/(4l).

44. 44 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

45. 45 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…) 12311 2 2323

46. 46 Proof of Lemma 6 Because of symmetry, it suffices to prove Lemma 6 for the pattern sequence P 1. Suppose to the contrary that P 1 contains a subsequence of type up- down-up(k ). Thus, there is a subsequence S = s 1, s 2, …, s 3k -2 of P 1 such that the integers s 1,s 2,…s k are pairwise different and that, for i=1,2,…,k, s i =s 2k -i =s (2k -2)+i. For simplicity of notation, suppose that s i =i (i=1,…,k ) and that S=P 1,[1,3k -2]. We obtain a contradiction by showing that k of the edges e 1,e 2,…,e 3k -2 are pairwise parallel. 3 3 3 3 3 3 3 3

47. 47 Definition 9 Let i,j I, and let dir(v i,v j ) denote the direction of the vector v i,v j. Then: (i) i j, if i<j and dir(v i,v j ) [dir(e k ), ] (ii) i j, if i<j and dir(v i,v j ) [, + dir(e 2k -1 )] (iii) i j, if i<j and dir(v i,v j ) [ + dir(e 2k -1 ), 2 ] U (0, dir(e k )) 123 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 i 1,i 2,..,i k of length k. We distinguish the corresponding three possible cases. 123123

48. 48 Definition 9 (cont.) If i 1 i 2 …. i k, then the edges e i1, e i2, …, e ik are pairwise parallel. Indeed, if 1 ≤ j < j’ ≤ k then the inequalities 0 ≤ dir(e ij ) < dir(e ij’ ) ≤ dir(e k ) ≤ dir(v ij,v ij’ ) < show that the edges e ij, e ij’ are parallel. 12311 3 vijvij vij’vij’ e ij e ij ’ Similarly, if i 1 i 2 … i k, then the edges e (2k -2)+i1, e (2k -2)+i2,…,e (2k -2)+ik are pairwise parallel, and if i 1 i 2 … i k then the edges e 2k –i, e 2k -i, …, e 2k -i are pairwise parallel. 22 333 33 3 3 3

49. 49 The End…