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Crossing Number and Applications

Crossing Number and Applications. Greg Aloupis (based on a seminar by Janos Pach and a journal paper by Tamal Dey). What’s a crossing number?. X(G) is the minimum number of edge crossings in any planar drawing of G. if X(G) = 0, then G is planar.

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Crossing Number and Applications

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  1. Crossing Number and Applications Greg Aloupis (based on a seminar by Janos Pach and a journal paper by Tamal Dey)

  2. What’s a crossing number? • X(G) is the minimum number of edge crossings in any planar drawing of G. • if X(G) = 0, then G is planar. • if X(G) = 1, then there are 41 minors that G does not contain (Robertson-Seymour ’93) • if X(G) = 2, we don’t know.

  3. Theorem: if e4v, X(G)  ke3/v2 • First by Ajtai-Chvatal-Newborn-Szemeredi in ’82, (k=1/100), and by Leighton ’83. • k has been raised over the years, but won’t exceed 8.

  4. Proof: if e4v, X(G)  e3/64v2 • Lemma: X(G)  e-(3v-6) > e-3v • Pick every vertex with probability p and obtain a subgraph G’. • Now, E[X(G’)] > E[e’] - 3E[v’] so p4X(G) > p2e - 3pv . X(G) is maximized when p4v/e.

  5. Applications • Number of incidences between n points and m lines is O(n+m+ n2/3 m2/3) • Szemeredi-Trotter ’83 • Number of unit distances formed by n points in the plane is O(n4/3) • Spencer-Szemeredi-Trotter ’84 • Number of distinct distances is cn4/5/logcn • Chung-Szemeredi-Trotter ’92

  6. Application IV: dividing lines Tamal K. Dey ’98

  7. Application IV: dividing lines

  8. Application IV: dividing lines

  9. Application IV: dividing lines

  10. Application IV: dividing lines

  11. Application IV: dividing lines • The result by Dey: there are O(nk1/3) k-sets for a planar set of n points. • In the dividing line case, k = n/2, so O(n4/3) . • Previous results: • Lovasz ’71 gave a bound of O(nk1/2). • Pach-Steiger-Szemeredi ’89 improved by a log*k factor…

  12. Suppose we have a planar graph with e edges corresponding to dividing lines of the n vertices. We want to prove that e is O(n4/3) .

  13. A really short “proof ” • Claim: the number of crossings, X, in such a graph is O(n2). • in general, O(nk) • We know that X(G)  e3/64n2 • And we assume e>4n, otherwise we’re done already. • O(n2) > X > X(G) > e3/64n2 • Combining, obtain a bound of O(n4/3) for e. • in general, O(nk1/3)

  14. Longer proof: Our n points in the plane

  15. We obtain n lines in dual plane

  16. Lines in the plane

  17. …will map to points in the dual

  18. Important above/below relation:

  19. More important: intersections

  20. Reminder: we’re looking for lines between points, that have half points above/below

  21. Equivalent to looking for “special” convex intersections on the median level in the dual

  22. How many special vertices are there on the median level? It can revisit lines, so ??? • So here’s another approach: • Form n/2 concave chains, starting at x= -, one for each line starting under the median level. • Move along lines from left to right, turning right when hitting the median level (must be at a convex vertex)

  23. The median level…

  24. The median level… and example of a chain

  25. Note: chains don’t overlap or share vertices. They cover all special vertices on ML and all intersections below, but don’t overlap or cross over ML.

  26. Remember our graph? • Consider two (dividing) edges that cross. • Their intersection corresponds to a bridge between two chains in the dual

  27. Remember our graph? • Consider two (dividing) edges that cross. • Their intersection corresponds to a bridge between two chains in the dual

  28. So, the number of bridges between concave chains in the dual is an upper bound on the the number of crossings, X, in our graph.

  29. Flash back • Claim: the number of crossings, X, in our graph is O(n2). • in general, O(nk) • We know that X(G)  e3/64n2 • O(n2) > X > X(G) > e3/64n2 • Combining, obtain a bound of O(n4/3) for e. • in general, O(nk1/3)

  30. The number of bridges between concave chains in the dual is an upper bound on the the number of crossings, X, in our graph. • DONE • Number of bridges is less than the number of intersections among the concave chains, which is O(n2). • In general O(nk), Alon-Gyori ’86. • Thus X < #bridges < #intersections < O(n2).

  31. DONE • Proved: the number of crossings, X, in our constructed graph is O(n2). • in general, O(nk) • We know that X(G)  e3/64n2 • O(n2) > X > X(G) > e3/64n2 • Combining, obtain a bound of O(n4/3) for e. • in general, O(nk1/3)

  32. Summary of proof • Given n points, and e dividing lines (segments) • Go to dual: we have n lines, e special intersection points, which are on the median level of the arrangement. • Form n/2 vertex disjoint concave chains that “skim” the median level. • Every intersection among e edges corresponds to a bridge between concave chains. • The number of bridges is at most the number of intersections in the arrangement below median level. • The number of such intersections is at most quadratic. • So e3/v2 < X(G) < X < bridges < intersections(e) < O(v2) • So e is O(n4/3).

  33. Summary of proof • Given n points, and e dividing lines • Go to dual: we have n lines, e special intersection points, which are on the median level of the arrangement. • Form n/2 vertex disjoint concave chains that “skim” the median level. • Every intersection among e edges corresponds to a bridge between concave chains. • The number of bridges is at most the number of intersections in the arrangement below median level. • The number of such intersections is at most quadratic. • So e3/v2 < X(G) < X < bridges < intersections(e) < O(v2) • So e is O(n4/3).

  34. Summary of proof • Given n points, and e dividing lines • Go to dual: we have n lines, e special intersection points, which are on the median level of the arrangement. • Form n/2 vertex disjoint concave chains that “skim” the median level. • Every intersection among e edges corresponds to a bridge between concave chains. • The number of bridges is at most the number of intersections in the arrangement below median level. • The number of such intersections is at most quadratic. • So e3/v2 < X(G) < X < bridges < intersections(e) < O(v2) • So e is O(n4/3).

  35. Summary of proof • Given n points, and e dividing lines • Go to dual: we have n lines, e special intersection points, which are on the median level of the arrangement. • Form n/2 vertex disjoint concave chains that “skim” the median level. • Every intersection among e edges corresponds to a bridge between concave chains. • The number of bridges is at most the number of intersections in the arrangement below median level. • The number of such intersections is at most quadratic. • So e3/v2 < X(G) < X < bridges < intersections(e) < O(v2) • So e is O(n4/3).

  36. Summary of proof • Given n points, and e dividing lines • Go to dual: we have n lines, e special intersection points, which are on the median level of the arrangement. • Form n/2 vertex disjoint concave chains that “skim” the median level. • Every intersection among e edges corresponds to a bridge between concave chains. • The number of bridges is at most the number of intersections in the arrangement below median level. • The number of such intersections is at most quadratic. • So e3/v2 < X(G) < X < bridges < intersections(e) < O(v2) • So e is O(n4/3).

  37. Summary of proof • Given n points, and e dividing lines • Go to dual: we have n lines, e special intersection points, which are on the median level of the arrangement. • Form n/2 vertex disjoint concave chains that “skim” the median level. • Every intersection among e edges corresponds to a bridge between concave chains. • The number of bridges is at most the number of intersections in the arrangement below median level. • The number of such intersections is at most quadratic. • So e3/v2 < X(G) < X < bridges < intersections(e) < O(v2) • So e is O(n4/3).

  38. Summary of proof • Given n points, and e dividing lines • Go to dual: we have n lines, e special intersection points, which are on the median level of the arrangement. • Form n/2 vertex disjoint concave chains that “skim” the median level. • Every intersection among e edges corresponds to a bridge between concave chains. • The number of bridges is at most the number of intersections in the arrangement below median level. • The number of such intersections is at most quadratic. • So e3/v2 < X(G) < X < bridges < intersections(e) < O(v2) • So e is O(n4/3).

  39. Summary of proof • Given n points, and e dividing lines • Go to dual: we have n lines, e special intersection points, which are on the median level of the arrangement. • Form n/2 vertex disjoint concave chains that “skim” the median level. • Every intersection among e edges corresponds to a bridge between concave chains. • The number of bridges is at most the number of intersections in the arrangement below median level. • The number of such intersections is at most quadratic. • So e3/v2 < X(G) < X < bridges < intersections(e) < O(v2) • So e is O(n4/3).

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