ordinary lines extraordinary lines? james joseph sylvester prove that it is not possible to arrange...
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ORDINARY LINES
EXTRAORDINARY LINES?
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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, March 1893
Educational Times, May 1893
H.J. Woodall, A.R.C.S.
A four-line solution
… containing two distinct flaws
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First proof: T.Gallai (1933)
L.M. Kelly’s proof:
starting line
starting point
new line
new pointnear
far
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Be wise: Generalize!
or
What iceberg
is the Sylvester-Gallai theorem a tip of?
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A
B
C
E
D
dist(A,B) = 1,
dist(A,C) = 2,
etc.
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a b
x y z
a bx y z
This can be taken for a definition of a line L(ab)
in an arbitrary metric space
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).
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Lines in metric spaces can be exotic
One line can hide another!
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a bx y z
A
B
C
E
D
L(AB) = {E,A,B,C}
L(AC) = {A,B,C}
One line can hide another!
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a bx y z
A
B
C
E
D
L(AB) = {E,A,B,C}
L(AC) = {A,B,C}
no line consists of all points
no line consists of two points
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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)
A
B
C
E
D
L(AB) = {E,A,B,C}
L(EA) = {D,E,A,B}
C(AB) = {A,B,C,D,E}
L(AC) = {A,B,C}
C(AC) = {A,B,C,D,E}
Observation: In metric subspaces of Euclidean spaces, C(ab) = L(ab).
If at first you don’t succeed, …
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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.
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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.
Lemma 2: If some three points are contained in no closure line, then some closure line consists of two points.
Definitions: simple edge = two points such that no third point is between them simple triangle = abc such that ab,bc,ca are simple edges
Easy observation
Two-step proof
Step 1: If some three points are contained in no closure line, then some simple triangle is contained in no closure line.
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 triples)
(minimize dist(a,b) + dist(b,c) - dist(a,c) over all such triangles; then closure line C(ac) equals {a,c})
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b
5 points
10 lines
5 points
6 lines
5 points, 5 lines
b
5 points, 1 line
nothing between these two
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Every set of n points in the plane
determines at least n distinct lines unless
all these n points lie on a single line.
near-pencil
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Every set of n points in the plane
determines at least n distinct lines unless
all these n points lie on a single line.
This is a corollary of the Sylvester-Gallai theorem
(Erdős 1943):
remove this point
apply induction hypothesis to the remaining n-1 points
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On a combinatorial problem, Indag. Math. 10 (1948), 421--423
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 thatevery 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.
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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?
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“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.
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.
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z
x
y
Manhattan distance
b
a
a bx y z
becomes
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With Manhattan distance, precisely seven closure lines
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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.
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”.
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a bx zy
degenerate Manhattan lines:
aa
b
b
x
x yy
z
z
typical Manhattan lines:
a
a b
b
x
x
yy
z
z
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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.
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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.
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.
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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.
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 (Laurent Beaudou, Adrian Bondy, Xiaomin Chen, Ehsan Chiniforooshan, Maria Chudnovsky, V.C., Nicolas Fraiman, Yori Zwols 2012):
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.
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bipartite not chordal not distance-hereditary
chordal not bipartite not distance-hereditary
distance-hereditary not bipartite not chordal
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Theorem (Pierre Aboulker, Xiaomin Chen, Guangda Huzhang, Rohan Kapadia, Cathryn Supko 2014 ):
In every metric space on n points,
there are at least (1/3)n1/2 distinct lines or else
some line consists of all these n points.