Abstract:We seek the maximum number of colors in an edge-coloring of the complete graph K n not having t edge-disjoint rainbow spanning subgraphs of specified types. Let c(n, t), m(n, t), and r(n, t) denote the answers when the spanning subgraphs are cycles, matchings, or trees, respectively. We prove c(n, t) = n−1 2 + t for n ≥ 8t − 1 and m(n, t) = n−2 2 + t for n > 4t + 10. We prove r(n, t) = n−2 2 + t for n > 2t + 6t − and r(n, t) = n 2 − t for n = 2t. We also provide constructions for the more general problem in wh… Show more
“…Very recently, Jia, Lu, and Zhang [15] considered the analogous problem when the host graph is the complete bipartite graphs K p,q . Using similar approaches in [14], they proved that r(K p,q , 1) = (p − 2)q + 1 + δ pq for p ≥ q ≥ 4 and r(K p,q , t)…”
Section: Introductionmentioning
confidence: 90%
“…Both of the results in [14] and [15] leave a gap of Θ( √ t) when the number of edge-disjoint rainbow spanning trees is closer to the maximum possible number of edge-disjoint spanning trees in K n and K p,q respectively.…”
Section: Introductionmentioning
confidence: 97%
“…Akbari and Alipour [1] showed that r(K n , 2) = n−2 2 + 2 for n ≥ 6. Jahanbekam and West [14] extended the investigations to arbitrary number of edge-disjoint rainbow spanning trees (along with the anti-Ramsey number of some other edge-disjoint rainbow spanning structures such as matchings and cycles). In particular, for t edge-disjoint rainbow spanning trees, they showed that r(K n , t)…”
Section: Introductionmentioning
confidence: 99%
“…and they [14] conjectured that r(K n , t) = n−2 2 + t whenever n ≥ 2t + 2 ≥ 6. This conjecture was recently settled by the first and the third author in [19].…”
An edge-colored graph G is called rainbow if every edge of G receives a different color. Given any host graph G, the anti-Ramsey number of t edge-disjoint rainbow spanning trees in G, denoted by r(G, t), is defined as the maximum number of colors in an edge-coloring of G containing no t edge-disjoint rainbow spanning trees.
“…Very recently, Jia, Lu, and Zhang [15] considered the analogous problem when the host graph is the complete bipartite graphs K p,q . Using similar approaches in [14], they proved that r(K p,q , 1) = (p − 2)q + 1 + δ pq for p ≥ q ≥ 4 and r(K p,q , t)…”
Section: Introductionmentioning
confidence: 90%
“…Both of the results in [14] and [15] leave a gap of Θ( √ t) when the number of edge-disjoint rainbow spanning trees is closer to the maximum possible number of edge-disjoint spanning trees in K n and K p,q respectively.…”
Section: Introductionmentioning
confidence: 97%
“…Akbari and Alipour [1] showed that r(K n , 2) = n−2 2 + 2 for n ≥ 6. Jahanbekam and West [14] extended the investigations to arbitrary number of edge-disjoint rainbow spanning trees (along with the anti-Ramsey number of some other edge-disjoint rainbow spanning structures such as matchings and cycles). In particular, for t edge-disjoint rainbow spanning trees, they showed that r(K n , t)…”
Section: Introductionmentioning
confidence: 99%
“…and they [14] conjectured that r(K n , t) = n−2 2 + t whenever n ≥ 2t + 2 ≥ 6. This conjecture was recently settled by the first and the third author in [19].…”
An edge-colored graph G is called rainbow if every edge of G receives a different color. Given any host graph G, the anti-Ramsey number of t edge-disjoint rainbow spanning trees in G, denoted by r(G, t), is defined as the maximum number of colors in an edge-coloring of G containing no t edge-disjoint rainbow spanning trees.
“…Early results considered the problem with F being a single graph: see [1] for a survey and [4] for the notable determination of AR(n, C k ). More recent work considers problems where F consists of spanning subgraphs of K n ; see [2] for a discussion of such problems involving spanning cycles, perfect matchings, and spanning trees.…”
Let s(n, t) be the maximum number of colors in an edge-coloring of the complete graph K n that has no rainbow spanning subgraph with diameter at most t. We prove s(n, t) = n−2 2 +1 for n, t ≥ 3, while s(n, 2) = n−2for n = 4 (and s(4, 2) = 2).
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