Popular fairy problems 4

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 Wenda
Hans 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#

There are two differences between initial position and that immediately before white promotion: wRg5->b7 and wKg3->g5. But the way from the first to the second is not so simple - white must unparalyse own rook and let her pass mine field of black rooks. For that wS makes switchback and wB round trip.









ser-s#12 (6+4)
Madrasi
Daniel Novomesky
feenschach 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#

1.Kd7 2.RHd6 3.Ke6 4.Kf5 5.RHe5 6.RHg5 7.Kf6 8.RHg6 9.Kg7 10.Kh8 11.RHg7 12.RHg8 Rh1#

Crystal clear far chameleon echo of model mate.









ser-h#12 (2+3)
0+2 rookhopper
Holger Helledie
2nd 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#

Equihopper is needed at d2 to allow bK's passage to e1 and then at f2 where he blocks. Beautiful geometry, equihopper visits all 16 squares reachable by him.









ser-h#18 (4+3)
equihopper h2
Andrej Frolkin
Sergej 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=

And six queen promotions! Construction is perfect - Qa4 blocks, Qb1 guards b8 and b7, Qd7 guards a7, Qh1 checks - but what about Qe2? It turns out there is no other square for her to be immobilised as f3, g4 and h5 are on used lines. Also Qd7 cannot guard a7 from d4 because of bBg7 line! Beautiful!









ser-s=26 (7+5)
Madrasi
Unto Heinonen
1st 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)#

AUW of four white pawns starting on 2nd rank. Nothing more should be said...









ser-s#35 (6+11)
Circe
Torsten Jönsson
2nd 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#

Very long march of black King. For mate to appear, it is necessary to move Bd8 to c1, move pg6 to g2 and move f4 away. But there are obstacles to that plan on the board, white moves pf4 to f2 and then captures it again to maintain the way to d2. Also pd7 must be captured on the way and it closes c1-g5 line, thus it must be captured. Nice logic.









ser-h#35 (12+1)
Circe
Viktor Syzonenko
5th 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#

Simple position, but the strategy employed by Black is not so simple. To let own Re1 go to f4... no, no, it's impossible. Black switches roles of rooks - Ra4 paralyses wR on file and Re1 goes to a4 and only then Re2 moves to f4. The finale is then clear and Madrasi-typical. Formal themes: switchback and round trip of bK, exchange of bK-bR places (moves 33-35).

+++ Composition In the Spotlight (CIS) No. 22 +++


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.









ser-h#35 (8+3)
Madrasi

Comments to Juraj Lörinc.
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