Transcript Ramsey type Problems in Random Graphs

Asymmetric Ramsey Properties
of Random Graphs involving Cliques
Reto Spöhel
Joint work with Martin Marciniszyn, Jozef Skokan,
and Angelika Steger
Ramsey Theory
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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.
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Question:
 How many people must attend the party so that the
assertion holds for ` > 3 people?
 Are these numbers finite?
Ramsey Theory
Ramsey (1930)
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Extensions:
 Color non-complete graphs (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 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).
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Problem: For any fixed graph F, integer k, and edge
probability p = p(n), determine
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Observation:
 The family of graphs satisfying R(F, k) is monotone
increasing.
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The property R(F, k) has a threshold (Bollobás,
Thomason, 1987).
Threshold for Ramsey properties
Łuczak, Ruciński, Voigt (1992) / Rödl, Ruciński (1993, 1995)
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Intuition:
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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 (!)
Asymmetric Ramsey properties
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What happens 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).
The 0-Statement
0-Statement
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Our proof is constructive:
 We propose an algorithm that computes a valid coloring
of Gn,p a.a.s.
 All previous proofs were non-constructive.
The coloring algorithm
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Algorithm proceeds in 2 phases:
1. remove edges from G one by one in some clever way
2. reinsert the edges 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
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Advanced Idea: In Phase 2, before coloring the inserted
edge, we may recolor an existing edge first.
Example: coloring without a red K4 nor a blue K5 (ctd.)
Sunflowers
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Sunflowers: `-cliques, each edge of which is
intersection with an r-clique.
Example: r = 5 and ` = 4
 Only `-cliques that are the
center of a sunflower contain
no edge that can be recolored.
 outer r-cliques may mutually
overlap!
 Def: `-clique is dangerous :=
`-clique is center of sunflower.
The coloring algorithm
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Basic idea: in Phase 1, successively remove edges which
are not exactly the intersection of an `-clique with an rclique.
 these can be colored directly when reinserted in Phase
2.
The coloring algorithm
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Advanced idea: in Phase 1, successively remove edges
which are not exactly the intersection of a dangerous `clique with an r-clique.
 these can be colored possibly after recoloring an
existing edge when reinserted in Phase 2.
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(recall: dangerous = center of sunflower = cannot guarantee
that there is an edge that can be recolored)
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
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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 a dangerous K3 and a Kr.
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.
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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 dangerous `-clique (center of a
sunflower).
Using this property iteratively for the edges in the
surrounding r-cliques (‘petals’ of the sunflower), we can
build a subgraph of G (‘sunflower patch’) which is either
too dense or too large to appear in Gn,p
Proof Sketch
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Example: r = 5 and ` = 4
Main technical difficulty: handle overlapping petals
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Deterministic Lemma: If
algorithm fails in Phase 1,
then G contains
 either a dense sunflower
patch (much overlap),
 or a large sunflower patch
(little overlap).
Probabilistic Lemma: With
high probability, Gn,p contains
neither a dense nor a large
sunflower patch.
Remarks
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The generalization to asymmetric Ramsey properties with
more than two colors is straightforward, since 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 sunflower petals
 determine graphs which may remain after Phase 1
Questions?