Here you go...a problem from 1867, courtesy of a Mr W. Greenwood.
The original problem stated mate in 4, but in 2013 , a mate in 3 is also there ( thanks to Houdini for confirming it :)
Enjoy !
White to play 2B5/5R1K/4p1p1/6kp/4P2R/6P1/8/8 w - - 0 1 |
Mate in 3 [ 1. Kg7 e5 2.Bf5 gxf 3. Rxf5# ]
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