Author: Dieter Buerssner
Date: 15:25:36 06/09/04
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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 2: a-i b-j c-k d-l e-m f-n g-h 3: a-j b-k c-l d-m e-n f-h g-i 4: a-k b-l c-m d-n e-h f-i g-j 5: a-l b-m c-n d-h e-i f-j g-k 6: a-m b-n c-h d-i e-j f-k g-l 7: a-n b-h c-i d-j e-k f-l g-m 8: ??? Now, how do you pair the 8th round? Regards, Dieter
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