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Miscellaneous Problems

Solution

Peter Wong
US Problem Bull., May 1988

[4k2r/P1p1p1pp/3pKp2/88/5p2/1P1PP3/8]

5+9. Wh. retracts one move and mates in 2

Retract -1. b6xBa7!!.

Then the bBa7 is a promotee and really is the bPa7 promoted on e1 or g1 after 6 captures. With four other captures by the bPf3 (from b7) and the wQB captured on c1, this accounts for all missing wh. units. Thus wPh2 had to promote so that it can reach a capture square, and it could not do this without visiting d7 or f7, disturbing the bK and breaking Black's right to O-O.

So that Black cannot castle. Mate in 2 is with 1. b6xa7!! and 2. a8=R/Q mate because the defense 1 ... O-O?? is not legal.