Computational Geometry Seminar

1. Computational GeometrySeminar Lecture 7 The “Crossing Lemma” and applications Ori Orenbach Crossing Lemma - Part I

2. What did we see till now ? Every planar graph with |V|=n vertices has at most 3n-6 edges (Euler’s formula) So, a graph with more than 3n-6 edges must have at least one crossing For Example K5: Ok, but how many crossings are there ? Can we minimize the crossings efficiently ? Crossing Lemma - Part I

3. Why is that important ? In real life crossings are very common. “The brick factory problem” (Turan) Minimizing crossings is important in many fields, for example VLSI chip area: Crossing Lemma - Part I

4. What will we see today? Crossing number definitions First lower bound on the crossing number Improved constant on the lower bound using the probability method (The crossing lemma) Tightness of the lower bound Improving the lower bound using pre-assumptions on the graph (bisection width) Crossing Lemma - Part I

5. Definitions Consider a simple graph G=(V,E) with n vertices and m edges (m>3n-6) We want to embed G into the plane (just as we did for planar graphs) Only now we know that we would have at least one crossing. The crossing number of G: cr(G) is the smallest number of crossings among all drawings of G. Crossing of more than two edges in one point is not allowed Crossing Lemma - Part I

6. Definitions In such a minimal drawing the following three situations are ruled out: No edge can cross itself Edges with a common end vertex cannot cross No two edges cross twice Crossing Lemma - Part I

7. Immediate lower bound Suppose that G is drawn in the plane with cr(G) crossings. Consider the following graph H: The vertices of H are those of G together with all crossing points The edges are all pieces of the original edges as we go along from crossing point to crossing point G H Crossing Lemma - Part I

8. Immediate lower bound Obviously H is a planar graph (and simple) |EH|= m+2cr(G) because every new vertex has a degree of 4. |VH|= n+ cr(G) And from Euler we get: m+2cr(G) ≤ 3(n+cr(G))-6 Cr(G) ≥ m-3n+6 (*) Crossing Lemma - Part I

9. Example Consider K6. we have that: Cr(K6)≥15-18+6=3 Indeed, we have a drawing with just 3 crossings: Crossing Lemma - Part I

10. Can we know cr(G) for every graph? The problem is believed to be NP complete So, we cannot calculate the crossing number of a graph G efficiently, but can we bound it efficiently? The bound proven before is good enough when m is linear in n, but not when m is larger compared to n For example if m≥4n then m-3n+6≥n+6 How can we improve the bound? Crossing Lemma - Part I

11. Theorem (Ajtai et al, Leighton, Chavatal, Newborn and others) Let K(n,m) denote the minimum number of crossing pairs of edges in a graph with n vertices and m edges, If m≥4n then K(n,m) Known as the “Crossing lemma” K(n,m) is the same as cr(G) Crossing Lemma - Part I

12. The Crossing Lemma - Proof We will prove by induction on n that: If n≤8 then G cannot have 4n edges, so it must be the case that n≥9 (notice that if n=9 then G=K9) Then, if (2) is proven, then (1) will follow immediately Crossing Lemma - Part I

13. The Crossing Lemma - Proof The induction base case: For n=9 we get and (2) follows from (1) We can see that (2) follows from (*) for every n≥10 and 4n≤m≤5n (Notice that the right side of (2) cannot exceed m-3n) Hence, We can assume that G is a graph with n≥10 vertices and m>5n edges and that (2) is valid for all graphs with fewer than n vertices. Crossing Lemma - Part I

14. The Crossing Lemma - Proof Denote by (G-x) the graph obtained from G by removing a vertex x and all connected edges) We notice that ((* Since G-x has at least m-(n-1) ≥5n-(n-1)>4(n-1) edges, we can use the induction on G-x, so we get: (mx is the number of edges in G-x) (**) We can show that (Why?) Crossing Lemma - Part I

15. The Crossing Lemma - Proof And now: (By (* (By (** (And Jensen (inequality Crossing Lemma - Part I

16. How can we improve the bound ? Improve the constant Improve the order of magnitude Can be done only by using some assumptions on the graph (will be shown later) Crossing Lemma - Part I

17. Improving the constant on the lower bound Many proofs given We will see a general proof using the “Probabilistic method *”, and an example in case that m≥4n (The probability method used to prove an existence of an object in a collection by showing that the probability it exists is positive) Crossing Lemma - Part I

18. The “Crossing Lemma” – An improved constant G=(V,E), |V|=n, |E|=m≥cn (for any c>3) Then we can show a lower bound on the number of crossings in G, such that: Crossing Lemma - Part I

19. Crossing Lemma (using probabilistic method) Consider a minimal drawing of G (containing minimal number of crossings) Let ‘p’ be a number between 0 and 1 (p will be chosen later) Generate a subgraph of G: H VH = Vertices of G chosen with probability p each EH = All the edges uv in G, that both u and v were chosen in VH (we get that any edge remains with probability p2) crH = Crossings in G that all four (distinct) vertices involved were chosen in VH (probability p4) Crossing Lemma - Part I

20. Crossing Lemma (using probability) Let np, mp crp be the random variables counting the number of vertices, edges and crossings in H Since cr(G) ≥m-3n+6>m-3n for any graph we have that: E(crp-mp+3np)≥0  E(crp)-E(mp)+E(3np)≥0  p4cr(G) –p2m +3pn≥0 * And we get that: *Note that p4cr(G) is the expected number of crossings inherited from G, but the expected number of crossings in H is even smaller Crossing Lemma - Part I

21. Crossing Lemma (using probability) Now, set p= cn/m (which is at most 1 by our assumption) And we get: Crossing Lemma - Part I

22. Crossing Lemma (using probability) For example if m≥4n then p=m/4n≤1 We can improve the constant by using different assumptions on the edges. Crossing Lemma - Part I

23. Improving the constant Theorem (Pach and Toth): let G be a simple graph drawn in the plane with cr(G) crossings, then First we will mention the following corollary: Corollary: The crossing number of any simple graph with at least 3 vertices satisfies: cr(G)≥5e(G)-25V(G)+50 Crossing Lemma - Part I

24. Improving the constant Corollary: The crossing number of any simple graph G with at least 3 vertices satisfies: cr(G)≥5e(G)-25V(G)+50 Proof Idea: A graph with e(G)≥ (k+3)(v(G)-2) must have one edge crossing at least (k+1) other edges, for 0≤ k≤4. By induction on e, we delete such an edge and the minimum number of crossings is: Crossing Lemma - Part I

25. Improving the constant Proof (Theorem) : Again, we will use the probabilistic method. We have that e≥7.5n>(4+3)(n-2) Using the corollary: cr(G)≥5e(G)-25V(G)+50 And we have that: p4cr(G)≥5p2m-25pn  *p=7.5n/m<=1 Crossing Lemma - Part I

26. Kt Kt Kt Tightness We want To show that the lower bound is tight (we cant do better than m3/n2) Consider n/t copies of Kt drawn on the plane: … n/t Crossing Lemma - Part I

27. Kt Kt Kt Tightness … Cr(Kt) = O(n4) m = t2*n/t=nt cr = t4*n/t = t3n = m3/n2 e(Kt) = O(n2) Crossing Lemma - Part I

28. Improving the order of magnitude The key is to have some pre assumptions on the graph, and use them to find a better bound Bisection width and crossing number Crossing number in graphs with monotone property Crossing Lemma - Part I

29. Definitions • A graph property P is said to be monotone if: • Every subgraph of a graph G with the property also satisfies P • Whenever G1 and G2 satisfies P, their disjoint union also satisfies P. Crossing Lemma - Part I

30. Definitions The bisection width of a graph b(G) is defined to be: The minimum is taken over all partitions of V(G) to 2 disjoint groups each n/3 Crossing Lemma - Part I

31. Crossing number and bisection width Theorem: let G be a graph with bounded degree, then: Proof: Consider a drawing of G with cr(G) crossings. Like before, we introduce a new vertex at each crossing, so we obtain a graph with n+cr(G) crossings Crossing Lemma - Part I

32. Crossing number and bisection width Proof continued: We obtain a graph H with n+cr(G) vertices We weight each new vertex with 0 weight and assign a weight of 1/n for old vertices, and we get by the planar separation theorem, that by deletion of at most: O((n+cr(G))1/2) vertices, H can be separated into H1 and H2 such that each one has at least n/3 elements and we get: b(G) ≤ O((n+cr(G))1/2) • Note that this is a revised version on the planar separation theorem using weights Crossing Lemma - Part I

33. Crossing number and monotone property For any monotone property P, let ex(n, P) denote the max numberof edges that a graph on n vertices can have if it satisfies P If the property P is “G does not contain a subgraph isomorphic to a fixed forbidden subgraph H”, we write: ex(n, H) for ex(n, P) Crossing Lemma - Part I

34. Crossing number and monotone property Theorem: Let P be a monotone property with ex(n, P)=O(n1+a) for some a>0 then there exists two constants c, c’>0 such that the crossing number of any graph G with property P, which has n vertices and e≥cnlog2n edges satisfies: Crossing Lemma - Part I

35. Crossing number and monotone property proof: We will use a slightly different version of the bisection width lower bound: Let G be a graph of n vertices, whose degrees are d1,…,dn Then: Will be used without a proof Crossing Lemma - Part I

36. Crossing number and monotone property Proof continued: Let P be a monotone graph property with ex(n, P) For some A,a>0 Let G be a graph with |V(G)|=n, |E(G)|=e Suppose G satisfies P, and e≥cnlog2n We assume by contradiction that: Crossing Lemma - Part I

37. Crossing number and monotone property Proof idea: We will use a decomposition algorithm: On every step i of the algorithm we will look at the Mi components of the graph: Gi1, …, GiMi On every step of the algorithm every Gij falls into 2 components, each at most (2/3)n(Gij) vertices We will stop when the number of vertices in each component is small enough Crossing Lemma - Part I

38. The “Decomposition algorithm” k steps Crossing Lemma - Part I

39. Crossing number and monotone property The algorithm: Step 0: let G0=G, G10=G, M0=1, m0=1 On every step of the algorithm we will look at the Mi components of the graph: Gi1, …, GiMi Each component has at most (2/3)in vertices We can assume (without loss of generality) that: The first mi components of Gi have at least (2/3)i+1n vertices and the remaining Mi-mi has fewer Crossing Lemma - Part I

40. Crossing number and monotone property Proof continued: We have that The stopping rule: Else: delete for j=1,…,mi b(Gji) edges from Gij such that Gij falls into 2 components, each at most (2/3)n(Gij) vertices Crossing Lemma - Part I

41. Crossing number and monotone property Let Gi+1 denote the resulting graph on the original set. Each component of Gi+1 has at most (2/3)i+1n vertices Suppose the decomposition algorithm terminates in step k+1, then if k>0: Crossing Lemma - Part I

42. Crossing number and monotone property Proof continued: Using that for any non-negative real number: We get: Crossing Lemma - Part I

43. Crossing number and monotone property Proof continued: Denoting the degree of a vertex v in Gij by d(v, Gij) We get: Crossing Lemma - Part I

44. Crossing number and monotone property Proof continued: Now we use previous theorem about b(G), and we get that the total number of edges deleted during the procedure is: (m=cnlog2n) Crossing Lemma - Part I

45. Crossing number and monotone property Proof continued: And finally we get that e(Gk)≥e/2 Next we will find an upper bound on e(Gk): The number of vertices of each connected component is: Crossing Lemma - Part I

46. Crossing number and monotone property Proof continued: But each Gkj has the property P, since it is monotone, so it follows that: And from this, the total number of edges in Gk is: A contradiction ! And the theorem is proved Crossing Lemma - Part I