Abstract
Let T = (V, A) be a (finite) tournament and k be a non negative integer. For every subset X of V is associated the subtournament T[X] = (X, A boolean AND (X x X)) of T, induced by X. The dual tournament of T, denoted by T*, is the tournament obtained from T by reversing all its arcs. The tournament T is self dual if it is isomorphic to its dual. T is {-k}-self dual if for each set X of k vertices, T[V \ X] is self dual. T is strongly self dual if each of its induced subtournaments is self dual. A subset I of V is an interval of T if for a, b is an element of I and for x is an element of V \ I, (a, x) is an element of A if and only if (b,x) is an element of A. For instance, empty set, V and {x}, where x is an element of V, are intervals of T called trivial intervals. T is indecomposable if all its intervals are trivial; otherwise, it is decomposable. A tournament T ', on the set V, is {-k}-hypomorphic to T if for each set X on k vertices, T[V \ X] and T '[V \ X] are isomorphic. The tournament T is {-k}-reconstructible if each tournament {k}-hypomorphic to T is isomorphic to it. Suppose that T is decomposable and vertical bar V vertical bar >= 9. In this paper, we begin by proving the equivalence between the {-3}-self duality and the strong self duality of T. Then we characterize each tournament {-3}-hypomorphic to T. As a consequence of this characterization, we prove that if there is no interval X of T such that T[X] is indecomposable and vertical bar V \ X vertical bar <= 2, then T is {-3}-reconstructible. Finally, we conclude by reducing the {-3}-reconstruction problem to the indecomposable case (between a tournament and its dual). In particular, we find and improve, in a less complicated way, the results of [6] found by Y. Boudabbous and A. Boussairi.