Asymmetric Ramsey Properties of Random Graphs involving Cliques

1. Asymmetric Ramsey Propertiesof 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.: AAAAA

2. 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 ` > 3people? • Are these numbers finite?

3. Ramsey Theory Ramsey (1930) • Extensions: • Color graphs other than cliques (e.g., random graphs). • Avoid some fixed graph Fother thanK`. • Avoid graph F1 in blue and F2 in red (asymmetric case). • Allow more colors.

4. Random Graphs • Binomial model Gn,p • n vertices • include each edge with probability p, independently of all other edges • We study the limiting probabilitythat the random graph Gn,p satisfies a given property P, where p=p(n). • It turns out that many properties have thresholds functionsp0(n) such that

5. Ramsey properties 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). 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).

6. Łuczak, Ruciński, Voigt (1992)/Rödl, Ruciński (1993, 1995) Threshold for Ramsey properties • Intuition: • above the threshold, there are more copies of Fin 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 (!)

7. Łuczak, Ruciński, Voigt (1992)/Rödl, Ruciński (1993, 1995) Threshold for Ramsey properties • 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!]

8. R(L, R) Denote the family of all graphs that contain either a red copy of graph Lor a blue copy of graph Rin every edge coloringwith red and blue byR(L, R). Asymmetric Ramsey properties • What happens is if we want to avoid differentgraphs Fi in different colors i, 1·i·k? • We focus on the case with two colors.

9. 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).

10. Proving the 0-Statement

11. The 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. 0-Statement

12. The coloring algorithm • Algorithm proceeds in 2 phases: • remove edges from G one by one in some clever way • 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

13. 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 ignorecliques 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.

14. Phase 1: Remove edges successively and build stack S containing • copies ofK` we don’t care about anymore (because they contain an edge which can be recolored), and • edges which are not in a copy ofK` 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. The coloring algorithm … … … • At the end of Phase 1, all edges and all copies ofK` are on the stack.

15. 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 • 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 1: Remove edges successively and build stack S containing • copies ofK` we don’t care about anymore, and • edges which are not in a copy ofK` we still care about

16. 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+1is 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

17. 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 • 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 disjointcopy 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

18. Proof Sketch • Sunflowers: copies of K`, each edge of which is intersection with a copy of Kr. • Example:r = 5and`=4 • 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. • 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 ofK` is center of a sunflower. • However:outer Kr’s may mutually overlap!

19. 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

20. About the 1-Statement

21. 1-Statement The 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)

22. Questions?