The American Joseph Wainwright (1851 – 1921) is the composer, known for his tasks with two-mover problems.
Each problem is producing the next one just after the key-move is played, while the number of moves for the solution is increased by 1. To be exact...
...the initial position is Mate in 2 moves,
after the key is Mate in 3 moves,
after the key is Mate in 4 moves,
after the key is Mate in 5 moves.
J. C. J. Wainwright,
American Chess Bulletin, 1910,
Mate in 2 moves.
(a) #2 (10 + 9),
(b) after the key of (a) #3,
(c) after the key of (b) #4,
(d) after the key of (c) #5
In the initial position Black is stalemated. The solutions are simple (with possible exception the five-mover) :
(a) 1.b4! (zugzwang) cxb4 2.Bxb4#
(b) 1.b5! (zz) axb5 2.cxb5 (zz) c4 3.Bb4#
(c) 1.b6! (zz) cxb6 2.Bxb6 a5 3.c7 a4 4.c8=S#
(d) 1.Kg5! cxb6 2.Bxb6 a5 3.c7
3...a4 4.c8=Q/B Ke5 5.Bc7#
3...Kxe6 4.c8=Q+ Kd6/Ke5 5.Bc7#
3...Ke5 4.c8=Q a4/Kd6 5.Bc7#
27-11-2009 : The friend reader Alotan has posted a comment :
Nice problem. The mate in 5 had many variations and I had to set it on the chessboard. The reason for comment, however, is that it reminded me a nice helpmate problem by Caillaud, with similar twinning mechanism :
First prize, Pitlochry TT 2003
(a) h#2 (5+2),
(b) Position of (a) before the mating move and h#2,
(c) Position of (b) before the mating move and h#2.
It is not exceptional or difficult, but it belongs to those problems that remain carved in the memory of the solver.
Dear readers, send the solution.