Transcript Asymmetric Ramsey Properties

Asymmetric Ramsey Properties
of 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
Ramsey (1930)

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.
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
Łuczak, Ruciński, Voigt (1992) / Rödl, Ruciński (1993, 1995)

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 (!)
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) and
F = K3, k = 2 (Friedgut et al., 2006, 100+ pages!)
[sort of: it remains open whether b0 = b0 (n) is a constant!]
Asymmetric Ramsey properties

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.
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).
Threshold for asymmetric Ramsey properties
Conjecture: Kohayakawa, Kreuter (1997)
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).
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 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.
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.
The coloring algorithm
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


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.
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.
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
Lemma
For p  bn-1/m2(K`, Kr), the Coloring Algorithm terminates
a.a.s. and produces a valid coloring of Gn,p.


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
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 Lemma:
K` that isIfnot
  Deterministic
algorithm
the center
fails
of ainsunflower
Phase 1,
then
G contains
contains
an edge that can be
recolored.

either a dense structure of

sunflowers
If algorithm(much
gets stuck
on
overlap),
G,orevery
copy
of K` is of
center

a large
structure
of sunflowers
a sunflower.(little overlap).
However: outer
Kr’s may
  Probabilistic
Lemma:
Both
mutually
events
areoverlap!
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?