László A. Székely
University of South Carolina
Supported in part by NSF DMS 071111
GraDR 2012 Crossing Number Workshop and Minischool
Valtice, Czech Republic, May 21, 2012
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Thickness
Skewness
Splitting number
Vertex deletion number
Page number
Genus
Crossing number
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“There were some kilns where the bricks were
made and some open storage yards where the
bricks were stored. All the kilns were connected
by rail with all storage yards. … the trouble was
only at crossings. The trucks generally jumped
the rails there and the bricks fell out of them; in
short this caused a lot of trouble and loss of time
… The idea occurred to me that this loss of time
could be minimized if the number of crossings of
the rails had been minimized. But what is the
minimum number of crossings?”
(P. Turán remembering in 1977)
ê n úê n -1úê m úê m -1ú
cr ( K n,m ) = ê úê
ê
ú
ú
ê
ú
ë 2 ûë 2 ûë 2 ûë 2 û
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Proofs: Zarankiewicz (1954) Urbanik (1955)
Gap found: Ringel and Kainen independently
Pach-Tóth (1998): “Which crossing number is
it anyway?” (Mohar 1995)
Imre Lakatos: “Proofs and Refutations”
applied Popper to mathematics analysing
Euler’s polyhedral formula and the concept of
real function
1 ê n úê n -1úê n - 2 úê n - 3ú
cr ( K n ) = ê úê
ú
ê
ú
ê
ú
4 ë 2 ûë 2 ûë 2 ûë 2 û
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Achieved in two distinct ways:
Soup can drawing
Throwing different slopes to different
hemispheres
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Crossing Lemma
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Bisection width lower bound
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Graph embedding
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Crossing Lemma (Leighton, Ajtai-ChvátalNewborn-Szemerédi): For a simple graph G
on n vertices and m edges, either m ≤ 4n or
m3
cr (G ) ³
.
2
64n
For a graph G on n vertices and m edges, with
edge multiplicity up to M, either m ≤ CMn or
m3
cr (G ) ³ C¢
.
2
Mn

If n<< m << n2
æ
n2 ö
min ç cr (G ) 3 ÷
G
m ø
è
converges to a constant (conjectured by Erdős
and Guy)
e ( A, B) = # of edges between A, B
b (G ) = min e ( A, B)
AÈB=V
n
A, B ³
3

Leighton (1982), Sýkora-Vrťo (1993), PachShahrokhi-Szegedy (1994)
16cr (G) + å d i2 ³ b(G)2 1.582
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G1=(V1,E1), G2=(V2,E2), |V1|≤|V2|
Embedding ω: G1−>G2: a pair of injections (φ,ψ),
where φ:V1−>V2 and ψ:E1−>(path set of G2) such
that ψ(uv) is a φ(u)−φ(v) path for all uv in E1
For e εE2, μω(e)=|{f εE1 : e εψ(f)}|
For u εV2, mω(u)=|{f εE1 : u εψ(f)}|
μω(e1)=μω(e2)=2
μω=maxe (μω(e))
φ
G2
G1
mω(u2) =3
μω=2
ψ
mω(u1)=mω(u3)=2
G2 with drawing
D(G2)
G1
ω

Two types of crossings occur
in the induced drawing of G1:
Induced
drawing
ID(G1)
◦ At crossing edges of G2
£ mw2 per crossings
æ m v ö
( ) ÷ at v
◦ At vertices of G2
£ç w
ç
÷
2
è
ø
1
2
cr (G1 ) £ cr (ID (G1 )) £ mw × cr (D (G2 )) + å mw2 ( v)
2 vÎV2
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Leighton (1982)
Shahrokhi-Sýkora-Sz-Vrťo (1994)
Assume G1 is embedded into G2 by ω.
Assume G2 is drawn as D(G2), inducing ID(G1).
1
2
cr (G1 ) £ cr (ID (G1 )) £ mw × cr (D (G2 )) + å mw2 ( v)
2
vÎV2
2
æ mw ( v) ö
cr (G1 ) 1
mw ( v)
cr (D (G2 )) ³
ç
÷
å
£ d ( v)
2
mw
2 vÎV2 è mw ø
mw
cr (G1 ) 1
2
cr (D (G2 )) ³
- å d ( v)
2
mw
2 vÎV2
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Let T(n,k,l,s) denote the minimum size of an luniform hypergraph on n vertices, such that
any k-element subset contains at least s edges
from the hypergraph.
T(n,k,l) = T(n,k,l,1)
Ringel observed that T(n,5,4) ≤ cr(Kn)
Analogously if s ≤ crg(Kp) then T(n,p,4,s) ≤ crg(Kn)
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If G has no k independent nodes T(n,k,2) ≤ e(G)
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If G, the complement, has no k-clique, then
æ n ö
e (G ) £ ç
÷ - T ( n, k, 2)
è 2 ø
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Counting method: count crossings in copies
of Kn in a drawn copy of Kn+1:
cr ( K n+1 )

n +1) cr ( K n )
cr ( K n ) cr ( K n+1 )
(
³
Û
£
n-3
And hence
cr ( K n )
$lim
n®¥ æ
n ö
ç
÷
è 4 ø
æ n ö
ç
÷
è 4 ø
æ n +1 ö
ç
÷
è 4 ø
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Katona-Nemetz-Simonovits (1964) on Turán
numbers
Improvement on a fixed-size problem
induces infinitely many improvements
deKlerk, Maharry, Pasechnik, Salazar, Richter
(2004) 83% of Zarankiewicz conjecture
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Pach, Spencer, Tóth (1999) improvement on
the Crossing Lemma (conjectured by
Simonovits): if G has girth >2r and m>4n, then
æ m r+2 ö
cr (G ) ³ Wç r+1 ÷.
èn ø
As m2>cr(G), G has at most
æ 1+ 1r ö
m = Oç n ÷
è
ø
edges tight for r=2,3,5.
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Pach, Spencer, Tóth (1999) used the bisection
width method to prove their girth theorem
Alternative proof through Crossing Lemma
and graph embedding (yielding explicit
constant):
Assume that G is drawn in its optimal
drawing (and for simplicity assume that G is
d-regular)
Define Gr by joining vertices of G if their
distance is r – they are joined by a unique rpath
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Draw Gr following closely the paths in the
drawing of G
nd )
(
C
r 3
n
2
£ cr (G ) £ cr (G ) r ( d -1)
r
2
first
category
2r-2
+ ( junk )
second
category
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The number of incidences between n points
and m lines on the plane is at most
(
c n + m + ( nm)

2
3
)
This is tight up to a constant multiplicative
factor
Number of incidences = 7
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The Erdős-Purdy Conjecture is true.
The number of incidences between n points
and m lines on the plane is at most
10
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60
(
n + m + ( nm)
2
3
)
Alternatively, if n½ ≤ m ≤ n2, then the number of
incidences is at most
10
60
( nm)
2
3
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Consider an “appropriate” “natural” graph
drawn in the plane.
Crossings, number of vertices or number of
edges of this graph should be related to the
quantity of interest.
Set lower and upper bound for the crossing
number of this graph
Analyze the results in terms of the quantity of
interest.
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Wlog every line has at least one point. The
graph G is already drawn in the plane:
◦ Vertices are the points
◦ Edges are the line segments connecting two
consecutive points on any one of the lines
◦ #of incidences on a line = #of edges on that line + 1
◦ |E(G)| = #of incidences − m
3
æ m ö
1 ( # incidences - m)
³
cr
G
³
ç
÷
( )
2
2
64
n
è
ø
or #incidences - m £ 4n
(
#incidences £ max m + 4n,32 ( mn) 3 + m
2
)
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For all ε > 0, there is a constant cε such that
the number of unit distances among n points
on the plane is at most cε n1+ε
Erdős’ construction shows that the number of
unit distances can be n1+(c/log log n) (which is
bigger than n logn, even bigger than n logkn).
d 2 = x 2 + y 2 , where x, y are nonnegative integers
2
n = 7,
2 = 12 +12
11 unit distances
52 = 52 + 0 2 = 32 + 4 2
2
145 = 12 2 +12 = 82 + 9 2
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cn3/2 Erdős 1946
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o(n3/2) Józsa, Szemerédi 1973
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n1.444… Beck, Spencer 1984
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cn4/3 Spencer, Szemerédi, Trotter 1984
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n points in the plane determine at most O(n4/3)
unit distances.
Proof by crossing number method: Draw unit
circles around the points and define a drawn
graph G. Vertices= the n points, edges
connect consecutive points on the circles.
m=2# unit distances.
æ n ö
m3
çè 2 ÷ø ³ cr ( G ) ³ 64n 2 .
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Among n points and n unit circles in the
plane, what is the maximum number of
incidences?
The maximum number of incidences has the
same magnitude of growth as the maximum
number of unit distances among n points in
the plane, as n goes to infinity.
n = 2,
3 incidences
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Erdős-Hickerson-Pach (1989)
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Valtr
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Just topology cannot prove the Erdős unit
distance conjecture.
G
G2
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Embedding approach: consider the distance 2
multigraph G2 from G above. For simplicity,
assume d unit circles passes through every
point.
nd )
(
³ cr ( G ) ³ C
2 3
d n + nd
2
2
4
2
4
d
n³
M
Mn 2
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M<d anyway, but M<<d would give
improvement
“Something else” is needed
What could it be?
E.g. Elekes-Simonovits-Szabó (2007):
3 points in the plane, n-n-n unit circles pass
through them. # of points covered by all 3
families =O(n2−ε).
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Elekes: n real numbers have at least n5/4
distinct sums or products
Pach, Sharir, and others: many more
incidence bounds (points, translates of
convex closed curves, etc.)
Andrews (Iosevits proof): For a convex
polygon in the plane with n lattice vertices
n=O(area1/3)
Dey: # of planar k-sets is at most 7n(k+2)1/3
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Erdős conjecture (1946): The number of
distinct distances among n points on the
n
plane is at least
c
log n
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Erdős’ matching construction is a square grid
of size n½×n½.
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Cn1/2 Erdős 1946
Cn2/3 Moser 1952
Cn5/7 Fan Chung 1984
Cn(58/81)−ε Beck 1984
n4/5/logcn Fan Chung, Szemerédi, Trotter 1992
Cn3/4 from a single point Clarkson, Edelsbrunner,
Guibas, Sharir, Welzl 1990
Cn4/5 from a single point Sz 1997
Cn6/7 Solymosi, Tóth 2001
Cn(4e/(5e-1))-o(1) Tardos 2002
Cn(19/22)-o(1) Katz 2003
Cn(48-14e)/(55-16e)-o(1) Katz, Tardos 2004
Cn/log n Nets, Katz 2010
Draw concentric circles around each point
with radiuses = distances from point
Very high edge-multiplicity is possible:
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Determine the quantity
g ( n) = min max ( A + A , A × A )
AÍR
A =n
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Clearly, g(n)=O(n2)
g(n)=Ω(n1+ε) Erdős-Szemerédi 1983
g(n)=Ω(n32/31) Nathanson
g(n)=Ω(n16/15) Ford
g(n)=Ω(n5/4) Elekes 1997
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Proof by Elekes:
P = ( A × A) ´ ( A + A) = {( ab, c + d ) : a, b, c, d Î A}
L = {{( ax, a¢ + x ) : x Î R} : a, a¢ Î A}
L =n
2
n £ P £ g ( n)
5
(
2
2
n £ # of incidences = O ( P L ) + P + L
3
2
3
)

Complex plane:
P Ì {( w1, w2 ) : wi Î C}
L Ì {{( w, aw + b) : w Î C} : a, b Î C}
(
# of incidences = O ( P L ) 10 + P + L
7
)
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Is there a proof not using Euler’s formula that
planar graphs drawn in straight line segments
have O(n) edges?
Yes – Pinchasi (2007)
Drawing triplet systems in C2:
Every triplet is contained by a complex line
Body of a triplet: convex hull in R4
Bodies may share vertex or boundary edge
How many triplets can be drawn? O(n)?
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Abstract simplicial complex: finite family of finite
sets, closed for taking subsets
Its dimension: max set size minus 1
1-dimensional abstract simplicial complex can be
identified with a graph
Embedding in Euclidean space: (d+1)-sets span a
d–dimensional simplex; simplices only overlap in
spans of common subsets
Embedded 1-dimensional simplicial complex:
rectilinear drawing of a graph
Theorem: every d–dimensional abstract simplicial
complex embeds in R2d+1, but some does not
embed in R2d.
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Two vertex sets: n points and m straight lines
Edges: incidences
Szemerédi-Trotter (1982)
|E(G)|=O(n+m+(nm)2/3)
Theorem: de Caen, S (1997)
3-path PcP’c’ # P3=O(nm)
Implies Szemerédi-Trotter through AtkinsonWatterson-Moran (1960)
s ( A)
£ s ( AAT A )
nm
3
v’
v
P
P’
v
v’
P
P’
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Is # C6=O(nm) in the incidence bipartite graph
of points and straight lines?
This would imply # P3=O(nm)
Slightly fails (by log log n factor for m=n) Klavík,
Kráľ, Mach (2011)
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Bisection width
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Embedding
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Convex crossing numbers
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Approximation algorithms
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Leighton (1982), Sýkora-Vrťo (1993), PachShahrokhi-Szegedy (1994)
16cr (G) + å d ³ b(G) 1.58
2
i
2
2
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Leighton (1982)
Shahrokhi-Sýkora-Sz-Vrťo (1994)
Assume G1 is embedded into G2 by ω.
Assume G2 is drawn as D(G2), inducing ID(G1).
1
2
cr (G1 ) £ cr (ID (G1 )) £ mw × cr (D (G2 )) + å mw2 ( v)
2
vÎV2
2
æ mw ( v) ö
cr (G1 ) 1
mw ( v)
cr (D (G2 )) ³
ç
÷
å
£ d ( v)
2
mw
2 vÎV2 è mw ø
mw
cr (G1 ) 1
2
cr (D (G2 )) ³
- å d ( v)
2
mw
2 vÎV2