Bondy is 70!!
November 17th, 2014
ORDINARY LINES
EXTRAORDINARY LINES?
Educational Times, March 1893
James Joseph Sylvester
Prove that it is not possible to arrange any finite number
of real points so that a right line through every two of them
shall pass through a third, unless they all lie in the same
right line.
Educational Times, May 1893
H.J. Woodall, A.R.C.S.
A four-line solution … containing two distinct flaws
First proof: T.Gallai (1933)
L.M. Kelly’s proof:
starting point
far
new point
near
starting line
new line
Sylvester-Gallai theorem:
Every n noncollinear points determine
at least one ordinary line.
Kelly and Moser (1958):
Every n noncollinear points determine
at least 3n/7 ordinary lines
as
long as n > 7.
Green and Tao (2012):
Every n noncollinear points determine
at least n/2 ordinary lines
as long as n is large enough.
Be wise: Generalize!
or
What iceberg
is the Sylvester-Gallai theorem a tip of?
y
x
a
z
b
Observation
Line ab consists of
all points x such that dist(x,a)+dist(a,b)=dist(x,b),
all points y such that dist(a,y)+dist(y,b)=dist(a,b),
all points z such that dist(a,b)+dist(b,z)=dist(a,z).
This can be taken for a definition of a line L(ab)
in an arbitrary metric space
x
a
y
b
z
Lines in metric spaces can be exotic
One line can hide another!
x
y
a
b
z
A
B
E
L(AB) = {E,A,B,C}
L(AC) = {A,B,C}
D
C
One line can hide another!
x
y
a
b
z
A
B
E
L(AB) = {E,A,B,C}
L(AC) = {A,B,C}
D
C
no line consists of all points
no line consists of two points
If at first you don’t succeed, …
Definition: closure line C(ab) is the smallest set S such that
* a and b belong to S,
* if u and v belong to S, then S contains line L(uv)
Observation: In metric subspaces of Euclidean spaces,
C(ab) = L(ab).
A
L(AB) = {E,A,B,C}
B
E
L(EA) = {D,E,A,B}
C(AB) = {A,B,C,D,E}
D
C
L(AC) = {A,B,C}
C(AC) = {A,B,C,D,E}
Conjecture (V.C. 1998)
Theorem (Xiaomin Chen 2003)
In every finite metric space,
some closure line consists of two points or else
some closure line consists of all the points.
The scheme of Xiaomin Chen’s proof
Lemma 1: If every three points are contained in some closure line,
then some closure line consists of all the points. Easy observation
Lemma 2: If some three points are contained in no closure line,
then some closure line consists of two points.
Two-step proof
Definitions:
simple edge = two points such that no third point is between them
simple triangle = abc such that ab,bc,ca are simple edges
Step 1: If some three points are contained in no closure line,
then some simple triangle is contained in no closure line.
(minimize dist(a,b) + dist(b,c) + dist(a,c) over all such triples)
Step 2: If some simple triangle is contained in no closure line,
then some closure line consists of two points.
(minimize dist(a,b) + dist(b,c) - dist(a,c) over all such triangles;
then closure line C(ac) equals {a,c})
How many distinct lines can be determined by five points?
10 lines
6 lines
b
5 lines
8 lines
b
1 line
b
How many distinct lines can be determined by n points?
Complete answer for all sufficiently large n:
Salamon and Erdos (1988).
Partial answer: Erdos (1943)
n lines
nothing between these two
1 line
b
Every set of n points in the plane
determines at least n distinct lines unless
all these n points lie on a single line.
(Erdős 1943): This is a corollary of the Sylvester-Gallai theorem
remove this point
apply induction hypothesis to the remaining n-1 points
Combinatorial generalization
Nicolaas de Bruijn
Paul Erdős
Let V be a finite set and let E be a family of of proper subsets of V such that
every two distinct points of V belong to precisely one member of E.
Then the size of E is at least the size of V. Furthermore, the size of E
equals the size of V if and only if E is either a near-pencil or else the family
of lines in a projective plane.
On a combinatorial problem, Indag. Math. 10 (1948), 421--423
Every set of n points in the plane
determines at least n distinct lines unless
all these n points lie on a single line.
What other icebergs
could this theorem be a tip of?
Question (Chen and C. 2006):
True or false? In every metric space on n points,
there are at least n distinct lines or else
some line consists of all these n points.
“Closure lines” in place of “lines” do not work here:
For arbitrarily large n, there are metric spaces on n points,
where there are precisely seven distinct closure lines
and none of them consist of all the n points.
Manhattan distance
x
a
y
b
z
z
becomes
y
a
x
b
With Manhattan distance,
precisely seven closure lines
Another conceivable generalization of Sylvester-Gallai:
In every finite metric space,
there is a uniquely defined line
or else some line consists of all the points.
Adrian’s counterexample:
all lines are of the type
or
True or false? In every metric space on n points,
there are at least n distinct lines or else
some line consists of all these n points.
Partial answer (Ida Kantor and Balász Patkós 2012 ):
Every nondegenerate set of n points in the plane
determines at least n distinct Manhattan lines or else
one of its Manhattan lines consists of all these n points.
“nondegenerate” means “no two points share their
x-coordinate or y-coordinate”.
x
z
y
a
b
typical Manhattan lines:
x
z
a
y
x
b
a
y
b
z
degenerate Manhattan lines:
z
x
b
y
a
b z
a
x
y
What if degenerate sets are allowed?
Theorem (Ida Kantor and Balász Patkós 2012 ):
Every set of n points in the plane
determines at least n/37 distinct Manhattan lines or else
one of its Manhattan lines consists of all these n points.
True or false? In every metric space on n points,
there are at least n distinct lines or else
some line consists of all these n points.
Another partial answer (C. 2012 ):
In every metric space on n points
where all distances are 0, 1, or 2,
there are at least n distinct lines or else
some line consists of all these n points.
Another partial answer (easy exercise):
In every metric space on n points
induced by a connected bipartite graph,
some line consists of all these n points.
Another partial answer (Laurent Beaudou, Adrian Bondy, Xiaomin Chen, Ehsan
Chiniforooshan, Maria Chudnovsky, V.C., Nicolas Fraiman, Yori Zwols 2012):
In every metric space on n points
induced by a connected chordal graph,
there are at least n distinct lines or else
some line consists of all these n points.
Another partial answer (Pierre Aboulker and Rohan Kapadia 2014):
In every metric space on n points
induced by a connected distance-hereditary graph,
there are at least n distinct lines or else
some line consists of all these n points.
bipartite
not chordal
not distance-hereditary
chordal
not bipartite
not distance-hereditary
distance-hereditary
not bipartite
not chordal
A week-long jam session in Montreal, June 2011
Laurent Beaudou
Maria Chudnovsky
Adrian Bondy
Vasek Chvatal
Xiaomin Chen
Nicolas Fraiman
Ehsan Chiniforooshan
Yori Zwols
Work in progress: Pierre Aboulker, Xiaomin Chen,
Guangda Huzhang, Rohan Kapadia, Cathryn Supko
Proof techniques
A
L(AB) = L(AE) = L(BE) = {A,B,C,D,E}
B
E
D
C
L(AC) = L(BC) = L(CE) = {A,B,C,E}
L(AD) = L(DE) = L(BD) = {A,B,D,E}
L(CD) = {C,D}
A
Equivalence relation
on the edge set of a
complete graph
B
E
D
C
We do not understand
the structure of these equivalence relations!
Is the problem of recognizing them in co-NP?
Adrian’s question:
What can you say about
the size f(n) of the smallest equivalence class
when no line consists of all n points?
HAPPY BIRTHDAY, ADRIAN !!!