Author: Madhavan
Date: 07:19:58 07/08/05
Go up one level in this thread
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,thats the initial step,and there you go on talking about openings in FRC,that's hilarious. :)
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