Author: Dieter Buerssner
Date: 13:01:53 06/10/04
Go up one level in this thread
On June 10, 2004 at 00:29:00, Vincent Diepeveen wrote: >On June 09, 2004 at 18:25:36, Dieter Buerssner wrote: > >>On June 09, 2004 at 08:23:00, José de Jesús García Ruvalcaba wrote: >> >>>It actually implies that for any number of players, we can find swiss pairings >>>up to n-3 rounds (you may assume n is even, if n is odd it is sharper and up to >>>n-2 rounds). Of course I am assuming no dropouts from the tournament and other >>>irregularities. >> >>Do I understand you correctly, that this is independent of already played >>rounds, and however they were paired. Let me give an example with 14 players >>(a-n) and 8 or more rounds. Say the first seven rounds were (color and results >>etc. ignored): >> >>1: a-h b-i c-j d-k e-l f-m g-n > 1-0 0-1 1-0 0-1 1-0 0-1 1-0 (which will be about first round result anyway) >>2: a-i b-j c-k d-l e-m f-n g-h >How can H play G in round 2? > >Please show a possible result in round 1 that makes it possible. I was not referring to actual Swiss pairing anymore. Only to Jose's theorem that is cited above. I udnerstood the sentence like this: However stupid you pair, you will always find pairings for at least n-3 rounds, that will not involve a repeated games. I was actually sure, that this cannot be true, and constructed a cheap example. >Swiss pairing is a very complex system. There are very big books about it and >many analysis done. Average book is hundreds of pages about it. > >Please show results with the pairings to proof *any* point. See the example from Peter's post with 6 players and 4 rounds (it was actually constructed by me). I am certainly no expert at Swiss pairing. I read the rules, and tried to build a scenario where exactly that happen ... Is there an error in the example shown by Peter (it includes all results, and it does take color, upfloat/downfloat rules etc. into account). I came to the example, while thinking about writing a Swiss pairing program. When I saw, that I cannot always follow the FIDE rules, even not the most important rule, that a game from a previous round must never be repeated, I gave up. My conclusion was, that the Swiss FIDE rules are not really consistent, and that some sort of look ahead would be needed. Graph theory mignt help, to allways find rather consistent pairings, but that would not follow the FIDE rules (and might almost be impossible without a computer program to do the pairings). I don't doubt, that in typical situations with many more participants than rounds, there will be no problem. However I seem to remember, that either in Paderborn or in some CCT event, Yace had to play 3 rounds with black. Regards, Dieter
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