For the 13th Sabra Composing Ty, we should compose an orthodox helpmate two-mover, where
B1 (first black move) : a black piece captures a white which closes a line
W1 (first white move) : a white piece, of line movement, arrives on this line
B2 (second black move) : the black piece, which made the capture, moves.
I made the following problem, with no distinction :
 (10 + 6) | Problem-601 Manolas Emmanuel original h#2 a) Diagram, b) bKe8→c8 a) 1.Bxf5+ Be4 2.Bxg6 Bxg6# b) 1.Rxa4 Ra1 2.Ra8 Rxa8# Bicolour Bristol, Orthogonal - Diagonal transformation, Black sacrifices, Duels wB with bB, wR with bR. |
The Exercise is the following :
For the following position, (heterodoxe, because on d8 there is a Grasshopper), there are solutions when the bK is placed on squares e8 and g8.
Try to find a very economical modification (translocation or addition or removal of a piece), with which the problem has solution when the bK is placed on square c8.
 3gk3/3p2p1/3P2P1/r4P2/2P5/7b/4KRB1/8 (7 + 6) (Grasshopper 0 + d8) | Problem-602a Manolas Emmanuel original h#2 a) Diagram, b) bKe8→c8, c) bKe8→g8 a) 1.Bxf5 Be4 2.Bxg6 Bxg6# b) 1.?? ?? 2.?? ?? c) 1.Rxf5 Rxf5 2.Gh8 Bd5# |
Solution of the exercise
The correct answer was send by the Slovakian composer Jaroslav Stun (visit his page here) :a) bRa5→b5, which is the most economical, and
b) +bPb6
Thank you chess-friend Jaroslav!
 3gk3/3p2p1/3P2P1/1r3P2/2P5/7b/4KRB1/8 (7 + 6) (Grasshopper 0 + d8) | Problem-602 Manolas Emmanuel original h#2 a) Diagram, b) bKe8→c8, c) bKe8→g8 a) 1.Bxf5 Be4 2.Bxg6 Bxg6# b) 1.Bxf5 Rxf5 2.Rb8 Rc5# c) 1.Rxf5 Rxf5 2.Gh8 Bd5# Bicolour Bristol, ODT, Black Quasi-sacrifices. (Not very good as a problem, due to repetition of moves). |
 3gk3/3p2p1/1p1P2P1/r4P2/2P5/7b/4KRB1/8 (7 + 7) (Grasshopper 0 + d8) | Problem-602b Manolas Emmanuel original h#2 a) Diagram, b) bKe8→c8, c) bKe8→g8 a) 1.Bxf5 Be4 2.Bxg6 Bxg6# b) 1.Rxf5 Rxf5 2.Ga5 Rf8# c) 1.Rxf5 Rxf5 2.Gh8 Bd5# |
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