6 research outputs found

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    Splitting a tournament into two subtournaments with given minimum outdegree

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    A {\it (k1,k2)(k_1,k_2)-outdegree-splitting} of a digraph DD is a partition (V1,V2)(V_1,V_2) of its vertex set such that D[V1]D[V_1] and D[V2]D[V_2] have minimum outdegree at least k1k_1 and k2k_2, respectively. We show that there exists a minimum function fTf_T such that every tournament of minimum outdegree at least fT(k1,k2)f_T(k_1,k_2) has a (k1,k2)(k_1,k_2)-outdegree-splitting, and fT(k1,k2)≤k12/2+3k1/2+k2+1f_T(k_1,k_2) \leq k_1^2/2+3k_1/2 +k_2+1. We also show a polynomial-time algorithm that finds a (k1,k2)(k_1,k_2)-outdegree-splitting of a tournament if one exists, and returns 'no' otherwise. We give better bound on fTf_T and faster algorithms when k1=1k_1=1.Un {\it (k1,k2)(k_1,k_2)-partage} d'un digraphe DD est une partition (V1,V2)(V_1,V_2) de son ensemble de sommets telle que D[V1]D[V_1] et D[V2]D[V_2] soient de degréß sortant minimum au moins k1k_1 et k2k_2, respectivement. Nous établissons l'existence d'une fonction (minimum) fTf_T telle que tout tournoi de degré sortant minimum au moins fT(k1,k2)f_T(k_1,k_2) a un (k1,k2)(k_1,k_2)-partage, et que fT(k1,k2)≤k12/2+3k1/2+k2+1f_T(k_1,k_2) \leq k_1^2/2+3k_1/2 +k_2+1. Nous donnons également un algorithme en temps polynomial qui trouve un (k1,k2)(k_1,k_2)-partage d'un tournoi s'il en existe un et renvoie 'non' sinon. Nous donnons de meilleures bornes sur fTf_T et des algorithmes plus rapides pour k1=1k_1=1

    Complementary Cycles Containing Prescribed Vertices in Tournaments

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    We prove that if T is a tournament on n vertices and x; y are distinct vertices of T with the property that T remains 2-connected if we delete the arc between x and y, then there exist disjoint 3-cycles C x ; C y such that x 2 V (C x ) and y 2 V (C y ). This is best possible in terms of the connectivity assumption. Using this we prove that under the same assumptions T also contains complementary cycles C 0 x ; C 0 y (i.e. V (C 0 x )[V (C 0 y ) = V (T ) and V (C 0 x ) " V (C 0 y ) = ;) such that x 2 V (C 0 x ) and y 2 V (C 0 y ) for every choice of distinct vertices x; y 2 V (T ). Again this is best possible in terms of the connectivity assumption. It is a trivial consequence of our result that one can decide in polynomial time whether a given tournament T with special vertices x; y contains disjoint cycles C x ; C y as above. This problem is NP-complete for general digraphs and furthermore there is no degree of strong connectivity which suffices to guarantee such cyc..
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