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. Lett, , vol.110, pp.621-624, 2010.

, Recall that a tournament is sparse if it admits an FAS which is a matching. In this section, 734 we show that MaxACT and MaxATT are polynomial-time solvable on sparse tournaments

, Note that packing vertex-disjoint triangles (and hence cycles) in sparse tournaments is 736 NP-complete, vol.9

, that is the 738 set of its backward arcs A(T ) is a matching. If a backward arc xy of T lies between two 739 consecutive vertices, then we can exchange the position of x and y in ?(T ) to obtain a sparse 740 tournament with fewer backward arc, Let T be a sparse tournament according to the ordering of its vertices ?(T )

, As 744 T is sparse it is clear that X 0 is a set of disjoint triangles. Moreover, it can easily be seen 745 that there exists an optimal packing of triangles (resp. cycles) of T which is the union of 746 an optimal packing of triangles, X 0 be the set of triangles made from a backward arc from B to A and the vertex x, p.747

, and build the 748 optimal solution for T . Therefore we can focus on the case where every vertex of T is the 749 beginning or the end of a backward arc A(T ), MaxATT or MaxACT on T we can solve the problem on

, A(T ) such that for any i ?

, t(e i )) is a triangle of T . Let ? be the 756 problem such that, given a digraph G = (V , A ), the objective is to find a maximum sized 757 subset of A such that the digraph induced by the arcs of the subset is a functional and 758 digon-free digraph, G be the digraph with vertex set V = {e i : i ? [b]} and arc set A defined 755 by: (e i e j ) ? A if (h(e i ), h(e j ), t(e i )) or (h(e i ), t(e j )

, t(e i )) if i < j and otherwise. Given a triangle ?(e i e j ), 762 let s(e j ) be the second vertex of ?(e i e j ); in other words, if ?(e i e j ) = (h(e i ), t(e j ), t(e i )), then 763 s(e j ) = t(e j ) and s(e j ) = h(e j ) otherwise. Informally, ?(e i e j ) corresponds to the triangle, vol.764

, Claim 19.1. Let X be a solution of ? (G ). The set X is an optimal solution if and only 767 if ?(X) is an optimal solution of

, We cannot have e i = e k as X induces 769 a functional digraph in G . Without loss of generality, we may assume that i < k, vol.770, p.609

S. Bessy, M. Bougeret, R. Krithika, A. Sahu, S. Saurabh et al., Zehavi 23:19 tournament. As h(e i ) < ? h(e k ) the arc h(e i )s(e j ) is not an arc of, Thus if, p.772

, But in this 773 case e i = e l and e j = e k , implying {e i e j

, Notice that the size of the solution does not change

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