Author: Robert Hyatt
Date: 14:25:00 06/16/99
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On June 16, 1999 at 16:32:55, Robert Pope wrote: >It has been mentioned that the Swiss Pairing System is designed to find an >absolute winner in a tournament, but if not enough rounds are played, the >ordering of the lower places is not obvious. > >Is it also true then that the pairing system also does the opposite at the same >time: finds the absolute worst-performing participant? I don't mean this >question as an insult to any of the participants. I was just wondering if this >is what the pairing system would do. > >Also, from my understanding, the pairings try to pair opponents with the same >number of points. What if most or all of the participants with the same point >score have already played each other? Can they be paired a second time? To answer your first question, it works like this. If you have N opponents, where N is a perfect power of 2, and you have log2(N) rounds, then you will know the best player, and the worst player (assuming no draws). But the others will be distributed from better to worse, but not precisely ordered. If you do log2(N)+1 rounds, you get the top 2 or 3 and the bottom 2 or 3, but the middle is still muddled. To get perfect ordering, you need N-1 rounds so that everybody plays everybody, and to improve that you need 2*(N-1) rounds where everyone plays everybody with both colors. Where this gets messy is if you have too many rounds. The rules for pairing avoid the same players meeting twice in one Swiss, so there are exception rules to handle this... Had accelerated pairings been used at the WCCC, this would have been a bigger problem. With 7 rounds it will be a problem before round 7 when the two top scores have already played and can't play again, so they get to play lower-ranked opponents with no easy way to 'break the tie' they have if they both win...
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