Author: Angrim
Date: 11:05:31 10/23/03
Go up one level in this thread
On October 22, 2003 at 15:50:50, Darren Rushton wrote: >[D] Bq1B1K2/3PpN2/P3Pp2/P1p2P2/2Pk1b1R/1p6/pN1P1P2/QR6 w - - 0 1 > >'If I could take one diagram to a desert island, it would be this one.' > >Leonid Yarosh - Mate in 4 >1st Prize Shakhmatyi v SSSR 1983 > >Here are the main variations of the solution recapitulated: > >1.a7! >1...axb1=Q 2.axb8=Q Qxb2(!) 3.Qxb3 Qc3 4.Qxc3 mate > Qe4(!) 3.Qxf4 Qxf4 4.Rxf4 mate >1...axb1=R(!) 2.axb8=R! Rxb2(!)3.Rxb3 Kxc4 4.Qa4 mate > 2.axb8=Q? Rxb2! 3.Qxb3 stalemate >1...axb1=B(!) 2.axb8=B! Be4(!) 3.Bxf4 B- 4.Be3(5) mate > 2.axb8=Q? Be4! 3.Qxf4 stalemate >1...axb1=N(!) 2.axb8=N! Nxd2(!) 3.Qc1! Ne4 4.Nc6 mate > 2.axb8=Q? Nxd2! 3.Qxf4+ Kc3 4.?? > 3.Rxf4+ Ne4 4.?? > 3.Qc1 Ne4 4.?? > >What do Fritz and friends make of this one? > >Regards, > >Daz I ran my proof number searcher on this, and it instantly(27k evals) found a mate in 9 that starts with Rxf4, shows that pn-search goes for the most-forcing mate rather than the shortest. I switched to alpha-beta search, and it took roughly 3 seconds to find the mate in 4 that starts with a7. Angrim
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