Original Fairy problems
JF-2022-I:
01.01.2022 - 30.06.2022
No. 1682 Andrew Buchanan &
James Malcom
Singapore / USA
original - 15.01.2022
white Rh1 Ke1 Pf2e2c3a3c4 Bb4 Sc5
black Kc1
#6 9+1
Rokagogo
Rokagogo
Solution: (click to show/hide)
In this #6, black king “is in a cage” and has only repetitive moves. So it is better to show the idea in ser-#n : White : Ké1 Rh1 Pé2 Black : Kç1 Ba1 ser-‡6 (3+2) C+ Rokagogo
1.é4 2.é5 3.é6 4.é7 5.é8=R 6.Ké3(Ré2)‡
But it is only a scheme, and I think it is possible to produce much better problems with it. Another weakness : in Rockagogo, the spectator is expecting more than one castling…
Hi Sebastien nice to hear from you hope you are well. Your points are well taken. In a way, the “Rokagogo” concept is a joke – this is really yet another Staugaard problem but (1) WinChloe can validate R but not S (2) it’s an attempt at a Staugaard directmate economy record, with the idea that people may think “Oh they are doing Rokagogo now”, well we aren’t heh.
Really the idea of an “economy record” can reduce artistic content, and I think that’s what’s perhaps happened here.
Dear Andrew, thank you, yes I am fine, no covid case for the moment ! 🙂
I think that the condition Rosue is more interesting to work than Rokagogo, because you can make castling with other pièces than Rook.
But in the case of the ser#6 I have proposed, with Roque 5.e8=Q is also possible !
Condition Roque…I want to say
The rokagogo definition doesn’t explicitly forbid castling to a square that is guarded, but not after the castling move (e.g. wKe1 wRh1 bMOe2).
Good point. This relates to an ongoing discussion in MatPlus. I guess guarded e1 (i.e. check) still prevents castling> what about guarded f1?