1 Introduction

The course is Probabilistic Combinatorics, being lectured by

JulianSahasrabudheday

We will be studying in particular the Ramsey numbers. Starting with the simplest to define:

R(k)=min{n:every red/blue colouring of E(Kn) contains a monochromatic Kn}.

Example. R(3)6: pick a vertex. Without loss of generality say it has at least 3 red neighbours. If any of these are connected by a red edge, then we get a red triangle by using the original vertex. If none of them are connected by a red edge, then we get a blue triangle between them.

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R(3)>5: by considering the following picture.

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Thus R(3)=6.

Main question: how fast does R(k) as k?

We will also study:

R(l,k)=min{n:every red/blue colouring of E(Kn) contains either a blue Kl or a red Kk}.

For a fixed graph H, we define

R(H)=min{n:every red/blue colouring of E(Kn) contains a monochromatic H}.

Example. R(k)=R(Kk).

For H a fixed graph, we define

ex(n,H)=max{e(G):G is a graph on n vertices and GH}.

Example. Mantel’s Theorem says that

ex(n,K3)=n24.

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1.1 Binomial Random Graph

Definition 1.1 (Binomial random graph). Given n, p(0,1), the binomial random graph G(n,p) is the probability space defined on all graphs on n vertices, where each edge is included independently with probability p.

1.2 Topics in this course

1.3 Brief introduction to R(k)

Theorem (Erdos-Szekeres, ’35). R(k)4k.

Proof sketch. Let n4k and let χ be a colouring of Kn. Pick a vertex v. It must have 12n neighbours connected to it by the same colour.

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Now ignore everything that was connected using the other colour. Pick a new vertex from what remains, and apply the process again:

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We continue until either A or B gets to size k. We basically just want

n(12)k(12)k1,

i.e. n4k.

How about a lower bound?

Example.

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This gives R(k)(k1)2+1. Quite a long way from 4k!

Theorem (Erdos, ’47). R(k)(1o(1))k2e2k2.

Notation. o(1) denotes a quantity that 0 as k.

Proof. Let n=(1o(1))k2eek2 we choose χ to be a random colouring of E(Kn) where the colour of each edge is chosen red / blue uniformly and independently. We have

(χ has a monochromatic Kk)=(K[n](k){K is a monochromatic clique})2nk2k22(enk)k2k2=(enk2(k12))<1

by the choice of n.

Big question: Is there an “explicit” construction that gives R(k)(1+𝜀)k, for some fixed 𝜀>0?