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nabla: It's not that hard to write an algorithm that can determine the winner regardless of the outcome of the outstanding matches most of the time isn't too hard:
For each player i, let W_i the number of wins he has, D_i the number of draws, and U_i the number of undecided games. There is a winner if there is an i such that W_i + D_i / 2 > W_j + D_j/2 + U_j, for all j != i. This is trivial to calculate. The only winners you might miss are the winners who finish with the same amount of points as other players, but win on SB.
(kaŝi) Se vi volas ŝpari rettrafikon, vi povas malpliigi la kvanton da informoj aperantaj en viaj paĝoj, en la Agorodoj. Provu ŝanĝi la nombron da ludoj en la Ĉefpaĝo kaj la nombron da mesaĝoj en ĉiu paĝo. (pauloaguia) (Montri ĉiujn konsilojn)