2020
DOI: 10.48550/arxiv.2001.05589
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Variations on twins in permutations

Abstract: Let π be a permutation of the set rns " t1, 2, . . . , nu. Two disjoint orderisomorphic subsequences of π are called twins. How long twins are contained in every permutation? The well known Erdős-Szekeres theorem implies that there is always a pair of twins of length Ωp ? nq. On the other hand, by a simple probabilistic argument Gawron proved that for every n ě 1 there exist permutations with all twins having length Opn 2{3 q.He conjectured that the latter bound is the correct size of the longest twins guarant… Show more

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Cited by 3 publications
(8 citation statements)
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“…A combinatorial application of this problem is studied in [9]. Similar questions can be asked for other combinatorial objects (largest disjoint isomorphic subtrees in a tree, largest disjoint order isomorphic pair of sub-permutations in a permutation -see [6] for applications of such problems).…”
Section: Isomorphic Graphsmentioning
confidence: 99%
“…A combinatorial application of this problem is studied in [9]. Similar questions can be asked for other combinatorial objects (largest disjoint isomorphic subtrees in a tree, largest disjoint order isomorphic pair of sub-permutations in a permutation -see [6] for applications of such problems).…”
Section: Isomorphic Graphsmentioning
confidence: 99%
“…Note that the containment of tight twins, and, in particular, order repetitions, is not monotone in the sense that the absence of tight twins of length k does not exclude the presence of tight twins longer than k. By using the probabilistic method, we proved in [4], Thm. 3.6, that there exist arbitrarily long permutations without tight twins longer than 12.…”
Section: §1 Introductionmentioning
confidence: 96%
“…Let tt prq pnq denote the largest integer k such that every permutation of length n contains tight r-twins of length k. Our result from [4], mentioned above, states that tt p2q pnq ď 12.…”
Section: §1 Introductionmentioning
confidence: 99%
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