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 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.
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.
Applications
I.
Number of incidences between n points
and m lines is O(n+m+ n2/3 m2/3)
•
II.
Szemeredi-Trotter ’83
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).