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.– 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.

Theorem: if e4v, 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.

Proof: if e4v, 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 p4v/e.

Applications

I. Number of incidences between n points and m lines is O(n+m+ n2/3 m2/3)

• Szemeredi-Trotter ’83

II. Number of unit distances formed by n points in the plane is O(n4/3)

• Spencer-Szemeredi-Trotter ’84

III. Number of distinct distances is cn4/5/logcn• Chung-Szemeredi-Trotter ’92

Application IV: dividing lines

Tamal K. Dey ’98

Application IV: dividing lines

Application IV: dividing lines

Application IV: dividing lines

Application IV: dividing lines

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…

– 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) .

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)

Longer proof: Our n points in the plane

We obtain n lines in dual plane

Lines in the plane

…will map to points in the dual

Important above/below relation:

More important: intersections

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

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

• 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)

• The median level…

• The median level… and example of a chain

– 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.

Remember our graph?

• Consider two (dividing) edges that cross.

• Their intersection corresponds to a bridge between two chains in the dual

Remember our graph?

• Consider two (dividing) edges that cross.

• Their intersection corresponds to a bridge between two chains in the dual

• 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.

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)

• 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).

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)

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).

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).

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).

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).

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).

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).

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).

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).