Asymmetric Ramsey Properties of Random Graphs involving Cliques Reto Spöhel Joint work with Martin Marciniszyn, Jozef Skokan, and Angelika Steger TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: A A A A A

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 non-complete graphs (e.g., random graphs). Avoid some fixed graph F other than K `. Avoid graph F 1 in blue and F 2 in red (asymmetric case). Allow more colors. Ramsey (1930)

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 G n, 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)

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 if we want to avoid different graphs F i 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)

The 0-Statement 0-Statement Our proof is constructive: We propose an algorithm that computes a valid coloring of G n, p a.a.s. All previous proofs were non-constructive.

The coloring algorithm Algorithm proceeds in 2 phases:  remove edges from G one by one in some clever way  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 K 4 nor a blue K 5

The coloring algorithm Advanced Idea: In Phase 2, before coloring the inserted edge, we may recolor an existing edge first. Example: coloring without a red K 4 nor a blue K 5 (ctd.)

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

The coloring algorithm 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. (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 G n, p in Phase 1 a.a.s. unless ` = 3.  Algorithm needs to be refined for triangles.

The trouble with triangles If p  bn -1/ m 2 ( K 3, K r ), then G = G n, p a.a.s. contains K r + 1. Algorithm gets stuck in Phase 1 since every edge of K r + 1 is the intersection of a dangerous K 3 and a K r. 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 G n, 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 G n, p Lemma For p  bn -1/ m 2 ( K `, K r ), the Coloring Algorithm terminates a.a.s. and produces a valid coloring of G n, p.

Proof Sketch Example: r = 5 and ` = 4 Main technical difficulty: handle overlapping petals 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, G n, p contains neither a dense nor a large sunflower patch.

Remarks The generalization to asymmetric Ramsey properties with more than two colors is straightforward, since the conjectured threshold depends only on the two densest graphs F i. 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?