Marianka 2017 - fairy solving competition

As you might know, Marianka Cup composing tourney allowing 30 days for composers was this year announced for any kind of problems featuring hoppers changing direction over the hurdle: moose, eagle, sparrow or hamster. The tourney was very successful with very good problems produced. But the choice of Marianka Cup has provided me an idea for fairy solving tourney too: to let solver try the turning hoppers as well. I added grasshopper as the best known fairy piece and voila!

There were 11 problems to be solved within time limit 100 minutes and each of them could bring 30 points to the solver.

The results were as follows:
1. Michel Caillaud - 293 points (of 330 possible)
2. Marek Kolcak - 115 points
3. Milan Svrcek - 30 points
4. Ladislav Packa - 0 points

Of the 11 problems solved during the competition, 4 were already present on the CCM:

- #2 with grasshoppers - the easiest twomover solved for 30 by everyone except Laco
- h#3 with grasshoppers - helpmate with 2 solutions with mates over immobilized hurdle - solved securely by winner
- s#2 with grasshoppers - well solved by Michel, who even has fallen for a while for one try, but then corrected himself
- ser-h#8 with grasshopper - Marek and Michel found both solutions for full points

The other problems from the competition are given below.
Stephen Emmerson
5th HM R. McWilliam MT 1999-2001

1.Bxe7! th. 2.Bg5#
1...Mf8 2.Bd6#
1...Mf5 2.Bc5#

Two Schifman defences are introduced by the key providing two flights e3 and f4, meaning that wB must make switchback from the battery set up by the key. Black thus activates halfpin on the g-file to a pin, expecting upinning interference. But besides selfpin, defences provide guard to one of flights, therefore White can use other battery checks guarding just one of flights provided by the key, avoiding unpin.

Only Michel solved this one (this information will be repeated a few times in varying ways).

#2 (10+5)
5+2 moose

Juraj Lörinc
Prize New Year Tourney
Thema Danicum 2003

1.Hec6? th. 2.He2#
1...Ga5+ 2.H4c5#
1...Hd3 2.Bc3#
1...Hc1 2.Hc2#

1.Hcc6! th. 2.Hc2#
1...Ga5+ 2.Hed5#
1...Hd3 2.Bc3#
1...He1 2.He2#
1...Gh6,Gxa2,Gd8 2.Qe1#
1...Rb5 2.Rxb5#

It is quite clear that White would like to open half-battery on the 4th rank (even more so if you notice unprovided check 1...Ga5+), but the question is which is the right way. Two possible keys to c6 provide mate for the check, so it is necessary to look for some refutation. 1...Hf2 is thus the most important move, otherwise there is nicely seen change between phases: mate change, mate transference and radical change to yield Z-23-55.

Both Marek and Michel found the right key.

#2 (11+7)
1+1 grasshopper, 2+2 hamster

Chris Feather
StrateGems 2010

1.Qe2 d8EA 2.EAe4 Qc3#

1.Kd2 dxc8EA 2.Se3 Qb2#

1.Kf4 dxe8EA 2.Qf3 Qd4#

It is not immediately clear how the eagle promotions could contribute to the checkmates if there is no relevant hurdle available for wQ support by wEA. But once you find one solution, two other offer themselves almost immediately. Te idea is quite fairy: wQ gives well known "orthodox" mate without support, but eagle prevents black pieces from entering square between wQ and bK thanks to repelling effect.

Solved only by Michel.

h#2 (4+10)
0+5 eagle

Yoshikazu Ueda
Problem Paradise 2008

1.Sf6 Mxf8 2.Rxf8 Mxg7#

1.f6 Mxc7+ 2.Kf7 Mxh8#

1.e6 Ma5 2.Se7 Md6#

1.d6 Mxg7+ 2.Kd7 Mxb8#

1.Sc6 Mexd8 2.Sxd8 Mxc7#

The main reason for selection of this helpmate without deep strategical analogy is the richness of checkmates on the offer against the almost the starting position. There is a bit of symmetry used, but also some specific mates, among them 2...Md6# with active pin is a star one.

Michel have found 3 solutions out of 5 for 18 points.

h#2 (3+16)
2+0 moose

Juraj Lörinc
5th Commendation Crete 2010

1.Rg5 Mh5 2.Rxb5 Kxb5 3.Kh3 Qa8 4.Rg4 Qh1#

1.Rc7 Md8 2.Rc3 bxc3 3.Kh4 Qb2 4.Rg5 Qh2#

1.Kf4 Mg4 2.Rc7 Mf7 3.Rc4 bxc4 4.Kf5 Qf3#

The official tourney of Crete congress requested h#3-n with white queen and at least one other white piece besides king and pawns. This turned out quite a difficult requirement, but it was possible to use also fairy pieces. This way the present diagram was born: how is it possible to get wQ out of the cage? White moose does not help much, sacrifices of black pieces help more, but in the meantime moose prepares the mating nets.

Marek have found one solution and Michel two. 1.Kf4 with two moose jumps was too difficult.

h#4 (8+5)
moose c1

Harald Grubert
5th Prize 65th TT feenschach 2013

a) 1.Ke2 2.Mf1 5.Kb5 6.Mca5 7.Ka6 8.Ma7 SPb6#

b) 1.Mg3 2.Kf4 3.Mdf5 4.Kg5 5.Mgh6 7.Kh7 8.Mh8 SPg7#

Nice echo (1,7) - twinning helps a lot, but I like especially the mating pictures where wSP gives mate over two different moose, so that it is not enough if one of them jumps away.

Michel full points.

ser-h#8 (2+3)
1+0 sparrow, 0+2 moose
b) c5 -» f6

Václav Kotesovec
Phénix 2011

1...EAb8 2.Ka5 EAc7 3.Ka6 Kb4=
1...EAc5 2.b5+ Kc3 3.a5 Kb2=

It is an interesting ability of eagle to pin a unit (b-pawn in this case) and at the same time use it as a hurdle to guard a square in the bK stalemate net.

It was an interesting case from the competition point of view. While Michel had everything fine, Marek found one solution correctly, but the other he found actually was not a solution, for a while he even thought he had found a cook, but no.

h=2,5 (2+3)
eagle d4

Comments to Juraj Lörinc.
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