On the KŁR conjecture in random graphs
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Citations
Independent sets in hypergraphs
Hypergraph containers
Combinatorial theorems in sparse random sets
Independent sets in hypergraphs
Extremal results in sparse pseudorandom graphs
References
The Probabilistic Method
On a Problem of Formal Logic
Regular partitions of graphs
On sets of integers containing k elements in arithmetic progression
Related Papers (5)
Frequently Asked Questions (14)
Q2. What is the rough idea of the conjecture?
The rough idea of the conjecture is that the probability that a graph is bad is so small that a simple union bound tells us that with high probability a random graph does not contain any bad graph – which implies that the authors may use the sparse embedding lemma the authors need.
Q3. What is the -regularity lemma for sparse graphs?
For sparse graphs – that is, graphs with n vertices and o(n2) edges – the regularity lemma stated in Theorem 1.1 is vacuous, since every equipartition into a bounded number of parts is ǫ-regular for n sufficiently large.
Q4. What is the probability that G′ does not satisfy S?
Since G′ ∈ Gℓ(H,n, dp, ǫ) and for each s ∈ [R], the graph G ′ s is simply a p′s-random subgraph of G ′, it follows from Chernoff’s inequality and the union bound that for fixed s ∈ [R], the probability that G′s 6∈ Gℓ(H,n, d ∗ps, ǫ) = Gℓ(H,n, dp · p ′ s/2, ǫ) is at moste(H) · 22n · exp(−ǫ2dpsn 2/16).
Q5. What is the condition that a partition of the vertex set of a graph is an?
The authors will say that a partition of the vertex set of a graph into t pieces V1, . . . , Vt is an equipartition if, for every 1 ≤ i, j ≤ t, the authors have the condition that ||Vi| − |Vj || ≤ 1.
Q6. What is the simplest way to make a graph triangle-free?
The triangle removal lemma of Ruzsa and Szemerédi [76] states that for every δ > 0 there exists an ǫ > 0 such that if G is any graph on n vertices that contains at most ǫn3 triangles, then G may be made triangle-free by removing at most δn2 edges.
Q7. What are the only edges of G?
These are the only edges of G. Let us also write G∗(H,n,m, p, ǫ) for the set of all G ∈ G(H,n,m, p, ǫ) that do not contain a canonical copy of H.Since the sparse regularity lemma yields graphs with varying densities between the various pairs of vertex sets, it may seem surprising that the authors are restricting attention to graphs where all the densities are equal (to m/n2).
Q8. What is the strength of the regularity lemma?
Often the strength of the regularity lemma lies in the fact that it may be combined with a counting or embedding lemma that tells us approximately how many copies of a particular subgraph∗Mathematical Institute, Oxford OX1 3LB, United Kingdom.
Q9. What is the probability of a random graph G obtained from H?
Consider the random graph G obtained from H by replacing each vertex of H by an independent set of size n and each edge of H by a random bipartite graph with pn2 edges.
Q10. What is the simplest way to show that each subgraph of G′′ induced by sets?
it suffices to show that each subgraph of G′′ induced by sets Vi1 , . . . , Vik , where i1, . . . , ik ∈ [t] form a copy Kk in R, contains a Kk-packing covering at least (1 − β)-fraction of its vertices.
Q11. What is the proof of the dense regularity lemma?
The proof of this theorem is essentially the same as the proof of the dense regularity lemma, with the upper uniformity used to ensure that the iteration terminates after a constant number of steps.
Q12. What is the general principle of the Kohayakawa-Rödl regularity method?
As the authors have already mentioned above, the usefulness of the dense regularity method relies on the existence of a corresponding counting lemma.
Q13. What is the proof of part (ii)?
Part (ii) is proved using the results of Conlon and Gowers [17] and hence hold with probability at least 1 −N−B for any fixed B > 0, provided that C and N are sufficiently large.
Q14. How do the authors show that every H-free subgraph G′ Gn,p?
The authors shall show that, conditioned on the above two events, every H-free subgraph G′ ⊆ Gn,p with δ(G ′) ≥ (1 − 33χ(H)−4 + γ)pn may be made (χ(H) − 1)-partite by removing at most γpn2 edges.