Kára-Pór-Wood conjecture

Big Line or Big Cliquein Planar Point Sets

Jozef Jirasekjirasekjozef@gmail.com

The Problem

• Given two integers k, l• Show that if we have “enough” points

in the plane, then there are either:

k points which can “see” each other

l points which all lie on a single line

or

There exists an integer N(k,l),such that for any n ≥ N,every arrangement of n points contains either:

enough

• l = 3 (three points on a line)– set N = k, from n ≥ N points pick any k.– if they can see each other, we are done.

– if two of them can not see each other,– we get a line with 3 points!

Simple cases

Simple cases

• k = 3 (3 points which see each other)– set N = l, let n ≥ N.– if all n points lie on a line, we are done.– otherwise, pick the smallest triangle.

• if two points can not see each other,• the triangle was not the smallest!

– therefore, three points of the smallest triangle must be able to see each other!

Proof by Induction?

• N(3, l) = l (pick the smallest triangle)

• For larger k:• Select N(k – 1, l) points• Find either:

– l points on a line, or– k – 1 points “seeing”

each other

• Find another point which “sees” all thek – 1 points

Will not work!

• Given k – 1 points which see each other, we can add an arbitrary number of points, such that:– no l points lie on a single line, and– no added point sees all the k – 1 points!

Known results

• Easy for k ≤ 3 or l ≤ 3 (as shown here).• Kára, Pór, Wood: k ≤ 4, all l.• Addario-Berry et al.: k = 5, l = 4.• Abel et al.: k = 5, all l.

Questions?

Ideas?

jirasekjozef@gmail.com