2013
DOI: 10.2478/s11533-013-0285-x
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On the minimal length of the longest trail in a fixed edge-density graph

Abstract: A nearly sharp lower bound on the length of the longest trail in a graph on vertices and average degree is given provided the graph is dense enough ( ≥ 12 5). MSC:05C35

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Cited by 2 publications
(4 citation statements)
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“…The assignment problem was formalized in the present work as the resolution of the longest trail problem [ 36 , 42 ]. To the best of our knowledge, this is the first time in the field of structural biology that this problem is formalized in a such way.…”
Section: Discussionmentioning
confidence: 99%
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“…The assignment problem was formalized in the present work as the resolution of the longest trail problem [ 36 , 42 ]. To the best of our knowledge, this is the first time in the field of structural biology that this problem is formalized in a such way.…”
Section: Discussionmentioning
confidence: 99%
“…Nevertheless, some results have been already obtained in the literature concerning the longest trail problem. A lower bound on the length of the longest trail was proposed in [36] for enough dense graphs. The author proves that for a connected graph G of n nodes such that its average degree is at least k ≥ 12.5, there exists a trail of length at least or n ≤⌈ k ⌉ + 2.…”
Section: Problem Formulationmentioning
confidence: 99%
“…For the lower bound, suppose that we are given a red/blue colouring of the complete graph Kn ${K}_{n}$. After removing, for each colour class, a suitable forest that meets all odd degree vertices (see, e.g., [3, Proposition 2.1]), we may assume that each colour class is Eulerian, in the sense that every vertex has even degree in both red and blue. However, this is not immediately helpful, since the colour classes may be disconnected.…”
mentioning
confidence: 99%
“…After deleting a bounded number of vertices, we may also assume that every vertex has red degree at least n2 $n\unicode{x02215}2$, which implies that the remaining graph is connected. By Szécsi [3, Theorem 1.3], which says that any connected graph with average degree t $t$ contains a trail of length )(t2+O(t) $\left(\genfrac{}{}{0.0pt}{}{t}{2}\right)+O(t)$, this then implies that there is a red trail of length at least 2n29+O(n) $2{n}^{2}\unicode{x02215}9+O(n)$, as required. Since the component containing this trail is Eulerian, we also have a red circuit of at least the same length.…”
mentioning
confidence: 99%