2 -
Fairy Chess Review, 1937
[12+10. Matt in zwei Zügen]
Assuming that black cannot castle, the solution is 1.Qxa4 ~ 2. Qa8#.
From the material balance and pawn structure it is evident that the Bd1 promoted at g8 from originally c2 pawn. Thus it has made 4 captures: c2/3xd3/4xe4/5xf5/6xg6/7. The remaining capture then has happened with b2xa3.
As no more wh. captures are available, black captures must account for pawns at h6 and h5. So black has captured the rooks with h7xg6xh5 and the bishop with e7xf6. Original f1,c8 bishops have been captured at home, of course.
The task then is to construct a position, where after the capture of the last black piece, there is as many black pawn moves available as possible in order to allow white to promote the bishop, retract it to d1 as well as put Kf5,Ne4 and Qc2 in place. Should this be possible without need to move K or KR the castling right will be preserved and there is no solution.
The optimum play starts from the position (or essentially similar pos.)
[4k2r/1p1p1p2/5p1P/3N1Kpp/p3NP2/P1p5/P1QPP1P1/3B4]
Note that either a5 or c5 has been necessary to get Ra8 and Qd8 out. Also b2xa3 has been necessary to get Bc1 and Ra1 out. Furthermore it is always possible to arrange tempos in the previous play so that it is white to move.
From the diagram position the game proceeds:
1. fxg7 hxg6 2. g8=B gxh5 3. Bh7 a4 4. Bc2 c6 5. Kf5 c5 6. Bf6 exf6 7. Bd1 c4 8. Qc2 c3 9. Ne4
and the problem position is reached. Unfortunately it is black to move and so it takes one move too long to restore white position. Hence black must have moved K or KR in the original position and cannot castle any more.
Solution by mr.mip