Author: Thomas Mayer
Date: 10:21:54 02/01/05
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Hi Peter, On February 01, 2005 at 13:09:56, Peter Fendrich wrote: >On February 01, 2005 at 13:01:56, Eiko Bleicher wrote: > >>On February 01, 2005 at 12:58:24, Uri Blass wrote: >> >>>On February 01, 2005 at 12:20:30, Thomas Mayer wrote: >>> >>>>Hi, >>>> >>>>just a hypotetic question: >>>>Let's think that the bishop side has several bishops, all of the same colour. >>>>The question is now: when not any of the bishops alone can stop the pawn is it >>>>possible to create a position where they all together can stop it anyway ? >>>>Or is it sufficent to say that it is a win when the lone pawn was checked in the >>>>TBs against each bishop and both checks say that it is won ? >>>> >>>>I hope you understand what I mean -> I can not really explain it to myself... :) >>>> >>>>a possible example: >>>> >>>>[D] 3K2b1/7P/8/8/8/8/b2k4/8 w - - 0 1 >>>> >>>>without the bishop on b2 this would be of course won and TBs give back a win >>>>when we check the pawn against both bishops... and of course according to five >>>>man this position is also won... the question is if it is possible to create a >>>>position with the above mentioned circomstances which is draw ?! >>>> >>>>Greets, Thomas >>>> >>>>P.S.: You see, I am working on Quarks endgame... :) >>> >>> >>>I am not sure if this is the example that you look for but 2 bishop can stop the >>>pawn by Be8 pin when one bishop cannot do it. >>> >>>Uri >>> >>>[D]k7/5P2/2b3K1/8/b7/8/8/8 b - - 0 1 >> >>Hello Uri, >> >>now taking the pawn to the f-file is cheating :) I've been trying on h and >>didn't find such cool things :) >Here is one. >[D]8/2K5/8/7P/8/8/6k1/7b w - - 0 1 >White will win but insert a bishop at f3 with draw. your examples are not what I did search... Look at Uri's example. In your case the f3 bishop would always draw even if the other one is missing. At the moment I think about examples where white must capture the bishop to win and can not when it is protected... hm... For what that is good for ? Well, at the moment I program internal node recognizers like Ernst Heinz described... And when it returns a result this have to be as correct as possible... else the search will get crazy... It's fun... :) Greets, Thomas
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