Author: Madhavan
Date: 21:49:51 07/08/05
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On July 08, 2005 at 15:43:05, Jorge Pichard wrote: >On July 08, 2005 at 10:19:58, Madhavan wrote: > >>On July 07, 2005 at 20:47:33, Jorge Pichard wrote: >> >>>Since the standard chess Opening is one of the 960 Fischer Random Positions, and >>>there are many Standard chess openings, could we simply determine the amount of >>>possible Chess960 openings by multiplying the amount of chess opening in the >>>standard chess position by 960, or the process in much more complicated than >>>that? >>> >>>Jorge >> >>you don't know what you are talking about >> >>how could you possibly make such an assumption?I don't know if you could even >>remember how to arrange all 960 different position, > >I don't want to waste time with you, but nobody has to know the initail 960 >positions, you simply have to download it and use it with any FRC engine. yeah,waste your time with FRC,you haven't got anyone's attention,get it? >initial 960 positions: http://www.chesstigers.de/download/chess960_regeln.pdf > >PS: That was not the question, and I don't think that you are capable of >answering it correctly anyway. Apparently you didn't understand the question. In the first place,you should be knowing how to arrange initial position.there you went on talking about amount of position in FRC by multiplying the amount of openings in standard,which is not true. >Jorge > >thats the initial step,and >>there you go on talking about openings in FRC,that's hilarious. >> >>:)
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