A Ramsey property of random regular and k‐out graphs

Abstract

In this study we consider a Ramsey property of random d‐regular graphs, G(n,d). Let rgoodbreakinfix≥2 be fixed. Then w.h.p. the edges of G(n,2r) can be colored such that every monochromatic component has order o(n). On the other hand, there exists a constant γgoodbreakinfix>0 such that w.h.p., every r‐coloring of the edges of G(n,2r+1) must contain a monochromatic cycle of length at least γn. We prove an analogous result for random k‐out graphs.

Large monochromatic components in expansive hypergraphs

“…Another possible direction for further study involves replacing large monochromatic components with long monochromatic paths. Letzer showed that in every 2-coloring of G(n, p) with p = ω (1) n , there is w.h.p., a monochromatic cycle (path) of order at least (2/3 − o(1))n. Bennett, DeBiasio, Dudek, and English [5] generalized this result showing that if p = ω (1) n k−1 , then a.a.s. there is a monochromatic loose-cycle (loose-path) of order at least ( 2k−2 2k−1 − o(1))n in every 2-coloring of H k (n, p).…”

“…For hypergraphs, Bennett, DeBiasio, Dudek, and English [5] proved that if G is an (r − 1)-uniform hypergraph on n vertices with e(G) ≥ (1 − o(1)) n r−1 , then mc r−1 (G) ≥ (1 − o(1))n and mc r (G) ≥ ( r−1 r − o(1))n. As for random graphs, it was independently determined in [2], [8] that with high probability 2 , mc r (G(n, p)) ≥ (1−o(1)) n r−1 provided p = ω (1) n , and it was determined (using the result mentioned in the previous paragraph) in [5] that mc r (H r (n, p)) ≥ (1 − o(1))n provided p = ω (1) n r−1 , and mc r (H r−1 (n, p)) ≥ (1 − o(1)) (r−1)n r provided p = ω(1) n r−2 . All of these results for random graphs use the sparse regularity lemma and thus only provide weak bounds on the error terms.…”

Large monochromatic components in expansive hypergraphs

“…which by the assumption implies that G has a monochromatic component of order at least t. Without loss of generality suppose this monochromatic component corresponds to C 1 1 and thus by the comments above, we have that C 1 1 has degree at least t.…”

“…Note that for 1 ≤ r ≤ k, mc r (G) = n = mc r (K k n ) if and only if the r-shadow of G is complete. 1 On the other hand, as first noted by Gyárfás and Sárközy [11], when r > k = 2 it is surprisingly possible for mc r (G) = mc r (K n ) provided G has large enough minimum degree. See [11], [17], and [10] for the best known results on this minimum degree threshold in the case k = 2, and [3] for a precise result on the minimum codegree threshold in the case r = k + 1 ≥ 4.…”

“…For hypergraphs, Bennett, DeBiasio, Dudek, and English [5] proved that if G is an (r − 1)uniform hypergraph on n vertices with e(G) ≥ (1 − o(1)) n r−1 , then mc r−1 (G) ≥ (1 − o( 1))n and mc r (G) ≥ ( r−1 r − o(1))n. As for random graphs, it was independently determined in [2], [8] that with high probability, 2 mc r (G(n, p)) ≥ (1 − o(1)) n r−1 provided p = ω (1) n , and it was determined (using the result mentioned in the previous paragraph) in [5] that mc r (H r (n, p)) ≥ (1 − o(1))n provided p = ω (1) n r−1 , and mc r (H r−1 (n, p)) ≥ (1 − o( 1)) (r−1)n r provided p = ω (1) n r−2 . All of these results for random graphs use the sparse regularity lemma and thus only provide weak bounds on the error terms.…”