Author: Howard Exner
Date: 21:16:27 01/24/98
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On January 24, 1998 at 23:43:57, Bruce Moreland wrote: > >On January 24, 1998 at 23:28:38, Dave Gomboc wrote: > >>Augh.. can anyone confirm that all I need to do to look these up >>correctly in my book is to add 938 to each of the problem numbers? > >Yeah, I think so. This is true for the first one, and it's true for the >last one, so it's probably true for the ones in the middle. > >The position at issue is #142 by my way of counting, and #1080 according >to the book. The positions are actually numbered both ways in the file >I use. > >b3r3/q2B2k1/3Q2p1/1p5p/3pP3/5P2/1p4PP/5RK1 b - - 0 1 > >The key here is 1. ... d3. > >Someone told me yesterday that this position appears in the BT2450 test >suite, a few plies later (I think that 1. ... d3 2. Kh1 are played), so >you are supposed to find 2. ... Qf2. > >If you could post the solution line for this, that would be great. I posted this on Amir Ban's original thread a while ago. Here it is again for you: After Qf2 white faultered with Qxd3? (not as good as the suggested Rd1 which I think both you and Ernst came up with).After Qxd3 the nice win begins with Bxe4 2.fxe4 Qxf1 3.Qxf1 Ra8 0-1.
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