Our 9th TT is dedicated to "popular fairy problems". The theme is very vaguely defined and thus the problems clearly "popular" for one composer may be very different from the compositions with that adjective in the view of some other problemist. That's why there is very wide scope for ideas. In 1st, 2nd and 3rd special example files we've seen some direct, help- and self-mates. Now let's look at some seriesmovers...

Klaus WendaHans Peter Rehm Schach-Echo 1987 |
1.Sg6 2.Kf4 3.Bf2 4.Bc5 5.Rd5 6.Rd7 7.Be7 8.Rb7 9.Bh4 10.Kg5 11.Sf4 12.a8R+ Kxb7# |

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ser-s#12 (6+4) Madrasi |

Daniel Novomeskyfeenschach 2001 |
1.RHd3 2.RHd5 3.RHc5 4.RHb5 5.RHc7 6.Kd7 7.RHe7 8.Kc7 9.RHb7 10.Kb8 11.Ka8 12.RHb8 Ra1# |

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ser-h#12 (2+3) 0+2 rookhopper |

Holger Helledie2nd Comm Feenschach 1972 |
1.EQb8 2.EQh4 3.EQb6 4.EQh6 5.EQb4 6.EQh8
7.EQb2 8.EQd2 9.Kg2 10.Kf2 11.Ke1 12.EQf8
13.EQd4 14.EQf6 15.EQd6 16.EQf4 17.EQd8 18.EQf2 Rc1# |

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ser-h#18 (4+3) equihopper h2 |

Andrej FrolkinSergej Komarov 10th HM Die Schwalbe 1991 |
1.h8Q 2.Qh7 3.Qb1 ... 5.h8Q 6.Qh5 7.Qe2 ... 11.h8Q 12.Qh4
13.Qa4 ... 18.h8Q 19.Qh3 20.Qd7 ... 25.h8Q 26.Qh1+ d1Q= |

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ser-s=26 (7+5) Madrasi |

Unto Heinonen1st Prize TT Excelsiors Circe/Madrasi 1996-2000 |
1.b4 ... 3.bxa6(Sg8) 4.axb7 5.b8B 6.Bg3 7.f4 8.Be1 9.Kg3 10.h4 ...
12.hxg6(pg7) 13.gxh7 14.hxg8S 15.Sxh6(Sb8) 16.Sxf5(pf7) 17.Se3
18.f5 ... 20.fxg7 21.g8Q 22.Qe8 23.Qxa4(pa7) 24.Qc6 25.a4 ...
27.axb6(pb7) 28.bxa7 29.axb8R 30.Rxb7 31.Rxf7 32.Rf1 33.Kf2
34.g3 35.Qg2 dxe3(Sg1)# |

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ser-s#35 (6+11) Circe |

Torsten Jönsson2nd Prize Springaren 1998 |
1.Kc3 ... 3.Ke1 ... 6.Kxf4(pf2) 7.Ke5 ... 9.Kc3 ... 12.Kxf2 ...
15.Ke5 ... 17.Kxd7(pd2) ... 20.Kf4 ... 24.Kxd2 ... 27.Ke5 ...
30.Kxd8(Bc1) ... 33.Kxg6(pg2) ... 35.Kh4 Sf5# |

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ser-h#35 (12+1) Circe |

Viktor Syzonenko5th Prize Schach-Echo 1981 |
1.Kf2 ... 7.Kd4 8.Rc4 9.Rc2 10.Ke5 ... 16.Kf1 17.Rce2 18.Rc1 19.Rc4 20.Kf2 21.Ke3 22.Rf2 23.Rf4
24.Kd4 25.Ra4 26.Kd5 ... 32.Kxa5 33.Kb4 34.Ra5 35.Ka4 gxf4# Spotlight comment by Juraj Lörinc:
The idea of the solution is oldfashioned at the first sight. Black has to put one rook to f4, the other on a5,
then own king to a5 and everything is done - checkmate gxf4#. But it is not possible outright as moves by bK
to e2 or e3 would be checks to wK. Therefore it is necessary to exchange roles of both black rooks - Ra4 goes to e2,
Re1 goes to c4, then Re2 goes to f4 and finally Rc4 to a5.
The problem demonstrates well the difference between switchback (as a multi-move manoeuvre) and round trip. Kf1-f2-f3-g4-g5-f6-e5-d5 is a way there, Kd5-e5-f6-g5-g4-f3-f2-f1 is a way back, using exactly the same squares and vectors in the opposite sense. For me this is just a switchback. On the other hand, other part of the bK movement is a round trip, namely moves 10.-15.Kd4-e5-...-f2, 21.Kf2-e3 and 24.Ke3-d4 form a closed path. There is intermezzo of 16. and 20. moves Kf2-f1-f2, but it is just a little switchback added to the round-trip. Note also the perfect white economy - even all white pawns remaining in the final position are needed for the mate, while Pa5 skillfully avoids other manoeuvre of bK and bR on the a-file. |

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ser-h#35 (8+3) Madrasi |

Comments to
Juraj Lörinc.

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