in reply to Re^5: Algorithm problem: Order matches by difference between players
in thread Algorithm problem: Order matches by difference between players

No, the Swiss system will actually optimize that people of equal strength play against each other as soon as possible - where equal strength is determined mostly by the performance in the tournament. That is, if I win the first three rounds, I will play against someone else who has won in the first three rounds (unless I'm the odd one out, in which I most likely play against someone with 2.5/3).
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Re^7: Algorithm problem: Order matches by difference between players
by davies (Monsignor) on Apr 19, 2011 at 06:25 UTC

    where equal strength is determined mostly by the performance in the tournament
    So, by pretending to the algorithm that every match is a draw before anything is played, the part of the logic I have quoted rates everyone equally. Then the second part of the Swiss algorithm matches players of equal results within the tournament to be highest rank against lowest rank - exactly what the OP wants. Unless I've misunderstood something.

    Regards,

    John Davies

      Eh, no. Swiss does order players by rank if they tie in score, but then splits the pile, and have the top of each pile play against each order. For instance, with 8 players, ranked from 1 to 8, the first round is pairs 1-5, 2-6, 3-7, 4-8.

      Also in Swiss, colour preferences are more important when pairing than ratings are.

      For details of the Swiss pairing rules, see FIDE Swiss Rules.

        I agree about colour, but only in Chess, although that can be treated as Home/Away for some sports. I've certainly played in some Swiss tournaments that work as I described. I'm not surprised that there are many different interpretations, so whatever the OP wants, he would be wise to check that the code does what he wants before relying on it.

        Regards,

        John Davies