Author: Peter Fendrich
Date: 10:09:56 02/01/05
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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. /Peter
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