Author: Vincent Diepeveen
Date: 11:05:50 02/01/05
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On February 01, 2005 at 14:02:18, Vincent Diepeveen wrote: >On February 01, 2005 at 12:20:30, Thomas Mayer wrote: > >You mean like this? > [D]8/8/P4b2/2K3k1/8/8/8/8 b - - 1-0 Whereas : [D]8/8/P4b2/2K3k1/8/8/8/b7 b - - Is a draw. And if you remove the f6 bishop it's a win for white again. > >>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... :)
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