2012
DOI: 10.37236/2338
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Tight Co-Degree Condition for Perfect Matchings in 4-Graphs

Abstract: We will give a tight minimum co-degree condition for a 4-uniform hypergraph to contain a perfect matching.

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Cited by 36 publications
(53 citation statements)
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“…Now assume that r = 3 and consider any U ⊆ V (H) \ V 0 with i P (U ) = (a 1 , a 2 , a 3 ) for some nonnegative integers a 1 + a 2 + a 3 = k + 1. Pick any (b 1 , b 2 , b 3 ) ∈ I µ P (H) and let c j = a j − b j for j ∈ [3]. Note that exactly one of the three consecutive integers Given a set S of k + 2 vertices, we call an edge e ∈ E(H) disjoint from S S-absorbing if there are two disjoint edges e 1 and e 2 in E(H) such that |e 1 ∩ S| = k − 2, |e 1 ∩ e| = 2, |e 2 ∩ S| = 4, and |e 2 ∩ e| = k − 4.…”
Section: 4mentioning
confidence: 99%
“…Now assume that r = 3 and consider any U ⊆ V (H) \ V 0 with i P (U ) = (a 1 , a 2 , a 3 ) for some nonnegative integers a 1 + a 2 + a 3 = k + 1. Pick any (b 1 , b 2 , b 3 ) ∈ I µ P (H) and let c j = a j − b j for j ∈ [3]. Note that exactly one of the three consecutive integers Given a set S of k + 2 vertices, we call an edge e ∈ E(H) disjoint from S S-absorbing if there are two disjoint edges e 1 and e 2 in E(H) such that |e 1 ∩ S| = k − 2, |e 1 ∩ e| = 2, |e 2 ∩ S| = 4, and |e 2 ∩ e| = k − 4.…”
Section: 4mentioning
confidence: 99%
“…Let H 2 be the k-graph whose edges are all k-sets that intersect S. Then H 2 does not contain a perfect matching and δ ℓ (H 2 ) = 1 − 1 − 1 k k−ℓ + o(1) n k−ℓ for all 1 ≤ ℓ ≤ k − 1. In recent years Conjecture 1.1 (and its exact counterpart) has received substantial attention [1,4,9,11,21,22,27,31,34,36,37,39,40,43,44,45]. In particular, the exact threshold is known for all ℓ such that 0.42k ≤ ℓ ≤ k − 1 as well as for a handful of other values of (k, ℓ).…”
Section: Introductionmentioning
confidence: 99%
“…Using the absorbing method, Rödl, Ruciński and Szemerédi [70,72] showed that (9) h k−1 (k, n) = n 2 + o(n) for all k ≥ 3. With long and involved arguments, they [74] were able to obtain an exact result when k = 3.…”
Section: Hamilton Cyclesmentioning
confidence: 99%
“…Assume that 1 ≤ l < k and k − l divides n. Since every (k − 1)-cycle of order n contains an l-cycle on the same vertices, (9) implies that h l (k, n) ≤ n 2 + o(n). On the other hand, it is not hard to see that h l (k, n) ≥ n 2 − k when k − l divides k. In fact, when k divides n, a Hamilton l-cycle contains a perfect matching thus Construction 1.1 provides this bound; when k does not divide n, this was proven by Markström and Ruciński [64, Proposition 2].…”
Section: Theorem 32 [74]mentioning
confidence: 99%
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