Author: Robert Hyatt
Date: 19:09:59 06/16/99
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On June 16, 1999 at 21:42:11, James T. Walker wrote: >On June 16, 1999 at 17:25:00, Robert Hyatt wrote: > >>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... > >Hello Bob, >The rules say if exactly two contestants end up tied its a single play off game. > That's easy enough. It gets a little messy if 3 or more end up tied though. >Jim Walker and if that game ends a draw? :) sum-of-opponent's scores is used.. and if that is equal, sum-of-opponents-opponent's scores are used... and that _can_ be equal as well... :) It can be interesting...
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