Author: Jim Monaghan
Date: 08:12:16 07/14/01
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On July 14, 2001 at 09:38:20, leonid wrote: >Hi! > >If you like to solve a mate then you have one position. > >[D]qr1b1rq1/P1P1P1P1/1q1k1q2/Q1RbR1Q1/1N1q1N2/1QnQnQ2/q1pKp1q1/3B4 w - - > >Please indicate your result. > >Thanks, >Leonid. Hi Leonid, You post interesting problems! But I think this one is not legal. One of the basic conventions in the problem world is for the postion to come from a game no matter how ludricous the game would have to have been. With using the retrogade analysis idea, I come up with the following: White has: 4P, 2N, 1B, 2R, 5Q, and 1K ... hence Black made one capture only and took a Whilte bishop (the one that moves on black squares). Black has: 2P, 2N, 2B, 2R, 6Q, and 1K ... hence White made one capture only and took a black pawn. White's 4 missing pawns became the extra queens and Black's 5 (uncaptured) pawns became his extra queens. Because only 2 captures were made (there are 30 units still on the board from the original 32), I can see how White c and e pawns can share the same files as the identical Black one's, that part is OK. But here is the kicker. I don't see how it could have been possible for the 4 White pawns and the 5 Black pawns to have passed each other to become queens when no other captures could have been made. Hence, this position is impossible. Sorry Lenoid. Is there any flaw in the above reasoning? Cheers, Jim
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