Bondy is 70!! November 17th, 2014

ORDINARY LINES EXTRAORDINARY LINES?

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

First proof: T.Gallai (1933) L.M. Kelly’s proof: starting line starting point new line new point near far

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?

a b x y z a b x 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).

Lines in metric spaces can be exotic One line can hide another!

a b x y z A B C E D L(AB) = {E,A,B,C} L(AC) = {A,B,C} One line can hide another!

a b x 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

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, …

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. 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})

b 10 lines 6 lines 5 lines b 1 line b 8 lines How many distinct lines can be determined by five points?

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 b 1 line nothing between these two

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

On a combinatorial problem, Indag. Math. 10 (1948), 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.

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?

“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.

z x y Manhattan distance b a a b x y z becomes

With Manhattan distance, precisely seven closure lines

In every finite metric space, there is a uniquely defined line or else some line consists of all the points. Another conceivable generalization of Sylvester-Gallai : 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”.

a b x z y degenerate Manhattan lines: a a b b x x y y z z typical Manhattan lines: a a b b x x y y z z

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 ): 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. 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.

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 Maria Chudnovsky Laurent Beaudou Adrian Bondy Xiaomin Chen Ehsan Chiniforooshan Nicolas Fraiman Yori Zwols Vasek Chvatal

Work in progress: Pierre Aboulker, Xiaomin Chen, Guangda Huzhang, Rohan Kapadia, Cathryn Supko

A B C E D A B C E D L(AB) = L(AE) = L(BE) = {A,B,C,D,E} L(CD) = {C,D} L(AC) = L(BC) = L(CE) = {A,B,C,E} L(AD) = L(DE) = L(BD) = {A,B,D,E} Equivalence relation on the edge set of a complete graph Proof techniques

Is the problem of recognizing them in co-NP? We do not understand the structure of these equivalence relations! 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 !!!