Family Communication about Genetics

Galvin, Kathleen M.

doi:10.1002/9781118540190.wbeic191

Cited by 1 publication

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“…) means: with ab the base of v we form R i by adding to R i−1 all triangles avw and bvw with (7) w ∈ X i−1 .…”

Section: Breadth First and Branchingmentioning

confidence: 99%

Tuza's Conjecture for random graphs

Park¹

2020

Preprint

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A celebrated conjecture of Zs. Tuza says that in any (finite) graph, the minimum size of a cover of triangles by edges is at most twice the maximum size of a set of edge-disjoint triangles. Resolving a recent question of Bennett, Dudek and Zerbib, we show that this is true for random graphs; more precisely: for any p = p(n), P(Gn,p satisfies Tuza's Conjecture) → 1 (as n → ∞). INTRODUCTIONIn this paper, we use matching and cover for triangle-matching and cover of triangles by edges, and ν and τ for the corresponding matching and cover numbers; thus, for a (finite) graph H, ν(H) is the maximum size of a set of edge-disjoint triangles and τ (H) is the minimum size of a set F of edges with the property that each triangle contains a member of F .We are interested in the celebrated Tuza's Conjecture: Conjecture 1.1 (Tuza [16]). For any graph H, τ (H) ≤ 2ν(H).The inequality is tight when H is K 4 or K 5 (or, e.g., a disjoint union of copies of these), and is not far from tight in other cases not related to these two examples (even if the graph is K 4 -free; see [9]). We will not survey the literature-see e.g. [9, 3]-and just mention that the best general result remains that of Haxell [8]:for every H, τ (H) ≤ 66 23 ν(H).Here we consider a question raised recently by Bennett, Dudek and Zerbib [3] and independently by Basit and Galvin [7]; informally: is Tuza's conjecture true for random graphs? More precisely, is it true that for any p = p(n) and G n,p the usual binomial (or "Erdős-Rényi") random graph, w.h.p. 1G n,p satisfies Tuza's Conjecture?In [3] this was shown to be true if p < c 1 n −1/2 or p > c 2 n −1/2 , with c 1 ≈ 0.48 and c 2 ≈ 4.25. (They work with G n,m , but, as usual, this is about the same as G n,p with p = m/ n 2 and we will stick to the binomial version.) Here we finish this story:(This is in some sense a failure: for a while it seemed to us that the gap in [3] might hide counterexamples to Tuza's Conjecture.) JK was supported by NSF Grants DMS1501962 and DMS1954035. 1 "with high probability," meaning with probability tending to 1 as n → ∞.

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“…) means: with ab the base of v we form R i by adding to R i−1 all triangles avw and bvw with (7) w ∈ X i−1 .…”

Section: Breadth First and Branchingmentioning

confidence: 99%