# No.1190 (CJF)

 No.1190 Chris Feather (England) Original Fairy problems JF – 2017(I): January – June

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 No.1190 Chris Feather England original – 25.02.2017 Solutions: (click to show/hide) white Kd1 Rh1 Bf7 Pe6g2g4 black Ka1 Sh3 Pg6h6 ser-h#23                                     (6+4) Alphabetic Chess 1.Ka1-b2 2.Kb2-c3 3.Kc3-d4 4.Kd4-e5 5.Ke5-f6 6.Kf6-g7 7.g6-g5 8.Kg7-h8 9.Sh3-f4 { } 10.Sf4*e6 11.Se6-f8 12.Sf8-h7 13.h6-h5 14.h5*g4 15.g4-g3 16.g5-g4 17.Kh8-g7 { } 18.Kg7-f6 19.Kf6-e5 20.Ke5-d4 21.Kd4-c3 22.Kc3-b2 23.Kb2-a1 Kd1-c2 # { (C+ by Popeye 4.75)} It is based on the well-known fact that in ABC a unit on h8 is doomed to stay there for ever... or is it?! (Author)

### 4 Responses to No.1190 (CJF)

1. Geoff Foster says:

Black has two tasks: 1) remove the wPe6 (which the bK can’t do), and 2) move the bS higher up the board. The bS must be given the move, so the bK moves all the way to h8. The bS captures the wPe6 and then moves to h7, so that it will be pinned after 14.hxg4. The bK returns to a1 along the long diagonal, completing a switchback. The wR does good work in pinning the bSh7 and also giving mate.

2. Luce Sebastien says:

The solution is really very pleasant to see !

3. Kjell Widlert says:

The problem is reminiscent of my own problem from Problemkiste, April 2014 (the solution was never published because the magazine ceased publication):
Kf2 Rb8 Bf7 Sb4 c8 Pa6 – Ka1 Ph7
sh#23 ABC
1.Kb2 7.Kh8 12.h1B 14.Bxa6 15.Bxc8 16.Kg7 17.Bh3 23.Ka1, Sc2#
(not 12.h1Q? because 16.Kg7?? would be illegal)
Same idea, different form.

4. seetharaman says:

That was nice Kjell Widlert. Your idea is superbly economical in a black minimal. The bishop promotion is neatly forced. Feather of course shows the theme without capture of white piece. Kudos!