Asymmetric Ramsey Propertiesof Random Graphs involving Cliques

Reto Spöhel

Joint work with Martin Marciniszyn, Jozef Skokan, and Angelika Steger

Ramsey Theory

Folklore Among every party of at least six people, there are at

least three, either all or none of whom know each other Equivalently: Every edge-coloring of the complete

graph on six vertices with two colors contains a triangle.

Question: How many people must attend the party so that the

assertion holds for ` > 3 people? Are these numbers finite?

Ramsey Theory

Extensions: Color graphs other than cliques (e.g., random graphs). Avoid some fixed graph F other than K`.

Avoid graph F1 in blue and F2 in red (asymmetric case). Allow more colors.

Ramsey (1930)

Random Graphs

Binomial model Gn,p

n vertices include each edge with probability p, independently of

all other edges We study the limiting probability

that the random graph Gn,p satisfies a given property P, where p = p(n).

It turns out that many properties have thresholds functions p0(n) such that

Ramsey properties

R(F, k)

Denote the family of all graphs that contain a monochromatic copy of graph F in every edge coloring with k colors by R(F, k).

Problem: For any fixed graph F, integer k, and edge probability p = p(n), determine

Observation: The family of graphs satisfying R(F, k) is monotone

increasing. The property R(F, k) has a threshold (Bollobás,

Thomason, 1987).

Threshold for Ramsey properties

Intuition: above the threshold, there are more copies of F in Gn,p than

edges. This forces the copies of F to overlap substantially and makes

coloring difficult. Order of magnitude of threshold does not depend on k (!)

Łuczak, Ruciński, Voigt (1992) / Rödl, Ruciński (1993, 1995)

Threshold for Ramsey properties

Łuczak, Ruciński, Voigt (1992) / Rödl, Ruciński (1993, 1995)

It is conjectured that the constants b and B can be replaced by b0(1§ ) for some b · b0 · B (“sharp threshold”).

verified for trees (Friedgut, Krivelevich, 2000) andF = K3, k = 2 (Friedgut et al., 2006, 100+ pages!)

[sort of: it remains open whether b0 = b0 (n) is a constant!]

Asymmetric Ramsey properties

R(L, R)

Denote the family of all graphs that contain either a red copy of graph L or a blue copy of graph R in every edge coloring with red and blue by R(L, R).

What happens is if we want to avoid different graphs Fi in different colors i, 1· i · k ?

We focus on the case with two colors.

Threshold for asymmetric Ramsey properties

Kohayakawa, Kreuter (1997)

The conjecture is true if L and R are cycles.

Marciniszyn, Skokan, S., Steger (RANDOM’06)

The 0-statement is true if L and R are cliques.The 1-statement is true if L and R are cliques (and KŁR-conjecture holds).

Conjecture: Kohayakawa, Kreuter (1997)

Proving the 0-Statement

The 0-Statement

0-Statement

Our proof is constructive: We propose an algorithm that computes a valid coloring

of Gn,p a.a.s. Our algorithm also covers the symmetric case (with

some exceptions) and the asymmetric case involving cycles.All previous proofs were non-constructive.

The coloring algorithm

Algorithm proceeds in 2 phases:

1. remove edges from G one by one in some clever way

2. insert the edges back in the reverse order, always maintaining a valid coloring

Basic Idea: edges which are not exactly the intersection of an `-clique with an r-clique can be colored directly when reinserted in Phase 2. successively remove such edges in Phase 1.

Example: coloring without a red K4 nor a blue K5

The coloring algorithm Advanced Idea: In Phase 2, we don’t need to color the

inserted edge directly, but may recolor an existing edge first. We can ignore cliques which contain an edge that can be recolored.

Example: coloring without a red K4 nor a blue K5 (ctd.) In fact, the recoloring of an existing edge

(from red to blue) can be postponed to later in Phase 2 by pushing the `-clique on a stack in Phase 1.

This includes the original idea: the edge we recolor later (from red to blue) may be the edge we just inserted.

The coloring algorithm

Phase 2: Work through S and if S.top is an edge:

insert and color red if S.top is a copy of K`:

check if all red, recolor one edge to blue if needed.

Phase 1: Remove edges successively and build stack S containing

copies of K` we don’t care about anymore (because they contain an edge which can be recolored), and

edges which are not in a copy of K` we still care about

At the end of Phase 1, all edges and all copies of K` are on the stack.

………

The coloring algorithm

If all edges are removed in Phase 1, Phase 2 produces a valid coloring:

No red copy of K` is created, because every copy of K` is examined in phase 2, and one edge is recolored to blue if needed.

No blue copy of Kr is created, because the recoloring never creates a blue copy of Kr.

The algorithm can be shown to remove all edges from Gn,p in Phase 1 a.a.s. unless ` = 3. Algorithm needs to be refined for triangles.

Phase 2: Work through S and if S.top is an edge:

insert and color red if S.top is a copy of K`:

check if all red, recolor one edge to blue if needed.

Phase 1: Remove edges successively and build stack S containing

copies of K` we don’t care about anymore, and

edges which are not in a copy of K` we still care about

The trouble with triangles

If p bn-1/m2(K3, Kr), then G = Gn,p a.a.s. contains Kr+1.

Algorithm gets stuck in Phase 1 since every edge of Kr+1 is the intersection of K3 and Kr (and there is no copy of K3 with an edge that can be recolored).

Even nastier substructures may appear. Solution:

Determine those structures. Color them separately at the beginning of phase 2.

Example: ` = 3, r = 4

Proof sketch

To do: Show that a.a.s all edges will be removed in Phase 1 (` > 3) or only easily colorable graphs remain (` = 3).

Proof idea: Graphs for which algorithm gets stuck contain substructures that do not appear in Gn,p.

If algorithm gets stuck on some graph G, every edge of G is contained in a copy of K`, every edge of which is contained in an otherwise disjoint copy of Kr,every edge of which ...

Use these properties to build a subgraph of G which is either too dense or too large to appear in Gn,p

Lemma

For p bn-1/m2(K`, Kr), the Coloring Algorithm terminates a.a.s. and produces a valid coloring of Gn,p.

Proof Sketch

Sunflowers: copies of K`, each edge of which is intersection with a copy of Kr.

Example: r = 5 and ` = 4 Every copy of K` that is not

the center of a sunflower contains an edge that can be recolored.

If algorithm gets stuck on G, every copy of K` is center of a sunflower.

However: outer Kr’s may mutually overlap!

Deterministic Lemma: If algorithm fails in Phase 1, then G contains

either a dense structure of sunflowers (much overlap),

or a large structure of sunflowers (little overlap).

Probabilistic Lemma: Both events are very unlikely to happen in Gn,p.

Remarks

The generalization to asymmetric Ramsey properties with more than two colors is straightforward, as the conjectured threshold depends only on the two densest graphs Fi.

It follows from known results that the proposed algorithm also works for the symmetric case (with some exceptions) and the asymmetric case involving cycles.

What about the general asymmetric case? It seems plausible that the proposed algorithm works in

most cases. Two main challenges:

deal with overlapping sunflowers determine graphs which may remain after Phase 1

About the 1-Statement

The 1-Statement

1-Statement

The proof is routinely obtained using Szemerédi’s Regularity Lemma for sparse graphs

(Kohayakawa, Rödl, 1997), and the KŁR-Conjecture (1997), a probabilistic embedding

lemma for (, p)-regular graphs.

A direct approach seems conceivable (Kohayakawa, Schacht)

Questions?