16 research outputs found

    A new sufficient condition for a Digraph to be Hamiltonian-A proof of Manoussakis Conjecture

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    Y. Manoussakis (J. Graph Theory 16, 1992, 51-59) proposed the following conjecture. \noindent\textbf{Conjecture}. {\it Let DD be a 2-strongly connected digraph of order nn such that for all distinct pairs of non-adjacent vertices xx, yy and ww, zz, we have d(x)+d(y)+d(w)+d(z)β‰₯4nβˆ’3d(x)+d(y)+d(w)+d(z)\geq 4n-3. Then DD is Hamiltonian.} In this paper, we confirm this conjecture. Moreover, we prove that if a digraph DD satisfies the conditions of this conjecture and has a pair of non-adjacent vertices {x,y}\{x,y\} such that d(x)+d(y)≀2nβˆ’4d(x)+d(y)\leq 2n-4, then DD contains cycles of all lengths 3,4,…,n3, 4, \ldots , n.Comment: 24 page

    A Note on Long non-Hamiltonian Cycles in One Class of Digraphs

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    Let DD be a strong digraph on nβ‰₯4n\geq 4 vertices. In [3, Discrete Applied Math., 95 (1999) 77-87)], J. Bang-Jensen, Y. Guo and A. Yeo proved the following theorem: if (*) d(x)+d(y)β‰₯2nβˆ’1d(x)+d(y)\geq 2n-1 and min{d+(x)+dβˆ’(y),dβˆ’(x)+d+(y)}β‰₯nβˆ’1min \{d^+(x)+ d^-(y),d^-(x)+ d^+(y)\}\geq n-1 for every pair of non-adjacent vertices x,yx, y with a common in-neighbour or a common out-neighbour, then DD is hamiltonian. In this note we show that: if DD is not directed cycle and satisfies the condition (*), then DD contains a cycle of length nβˆ’1n-1 or nβˆ’2n-2.Comment: 7 pages. arXiv admin note: substantial text overlap with arXiv:1207.564

    A sufficient condition for a balanced bipartite digraph to be hamiltonian

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    We describe a new type of sufficient condition for a balanced bipartite digraph to be hamiltonian. Let DD be a balanced bipartite digraph and x,yx,y be distinct vertices in DD. {x,y}\{x, y\} dominates a vertex zz if xβ†’zx\rightarrow z and yβ†’zy\rightarrow z; in this case, we call the pair {x,y}\{x, y\} dominating. In this paper, we prove that a strong balanced bipartite digraph DD on 2a2a vertices contains a hamiltonian cycle if, for every dominating pair of vertices {x,y}\{x, y\}, either d(x)β‰₯2aβˆ’1d(x)\ge 2a-1 and d(y)β‰₯a+1d(y)\ge a+1 or d(x)β‰₯a+1d(x)\ge a+1 and d(y)β‰₯2aβˆ’1d(y)\ge 2a-1. The lower bound in the result is sharp.Comment: 12 pages, 3 figure
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