Announcement of TT Marianka 2015 - Fairies C 22.8.2015

Theme: Any problem with aim of check (instead of usual checkmate). Any stipulations with check aim are allowed, as well as any fairy conditions and/or fairy pieces. Any twins, zero-positions, duplex, multiple solutions, tries or set play are allowed.

Judge: Juraj Lörinc.

Example: see the following problem already shown on CCM - r+2 with fairy pieces - or a few more problems below.

Submissions: personally in Marianka to Juraj Lörinc or by e-mail to

Deadline: 22.8.2015, 18:00 for both personal submissions in Marianka and e-mail submissions.

(slovenská verzia vypísania)

Téma: Akákoľvek úloha, v ktorej je cieľom hry šach (namiesto bežného matu). Sú povolené akékoľvek výzvy s cieľom šach, exopodmienky a exokamene. Akékoľvek dvojníky, nulové pozície, duplex, viac riešení, pokusy, zvodnosti alebo zdanlivé hry sú dovolené.

Rozhodca: Juraj Lörinc.

Príklady: pozrite nasledujúcu úlohu už skôr zverejnenú na CCM - r+2 s exokameňmi - alebo ďalšie príklady nižšie.

Príspevky: osobne v Marianke Jurajovi Lörincovi alebo e-mailom na

Uzávierka: 22.8.2015, 18:00 pre osobne podané súťažiace úlohy i pre e-mailové príspevky.

Alfred Karlström
Magasinet 1946

1.Kf2! th. 2.e4+
1...Ke4 2.d3+
1...Kf4 2.e3+
1...Kg4 2.h3+
1...Ke5 2.d4+
1...Kg5 2.h4+
1...Ke6 2.d8S+
1...Kf6 2.dxe8S+
1...Kg6 2.h8S+

The key gives two flights for a whole set of eight flights. In #2 this would be a very good key, but here the main point is rather in the motivation of the key than this formal element. White king has to provide enough space for bK as well as to avoid being checked himself.

+2 (6+5)

Holger Helledie
3rd Commendation feenschach 1973

1.Nf5 2.Nb3 3.Na5 4.Nc6 5.Nb8 6.Nh5 7.Nb2 axb2+

The destination square of nightrider is clear. However the way there is far from trivial. White must avoid squares where he would be forced to check the bK. The fact that the position has no pieces except nightrider kings and blocked pawns constellation, i.e. no technical pieces, is amazing.

ser-r+7 (3+2)
nightrider h1

Milos Tomasevic
Markus Ott

Prize Mat 1981

11.Kxg5 29.Kxe1 48.Kxh4 70.Kxh2 92.Kxg4 113.Kxg2 114.Kf3 115.Bg2 116.Bf1 117.Bxd3 124.Ka6 125.Bb5+ Kd5,Kc5+

White king must first eat its way to f3 so that Bh1 can move away from the corner. Then the black royal battery is prepared and fired. Just like the previous seriesmover of Holger Helledie, there is serious question how to multiply the main content. Of course, authors of the seriesmover in 125 moves were mostly interested in length.

ser-s+125 (2+16)

Bent Martin Muus
Problemkiste 1989

1.g8Q 2.Qd8 3.Qc7 4.e8R 5.Red8 6.Ke7 7.e6 11.Kd3 c4+

Obviously, wK cannot leave diagonal c8-h3 immediately and wB cannot leave it either. That is why White has to immobilize its Rd7 first and then wK can march to d3. Two promotions by the way, nicely done.

ser-r+11 (8+2)

Michel Olausson
5th HM Springaren 1991

1.axb7? th. 2.b8S+
1...Kxb7 2.Bc7+
1...Kb5 2.b8G+
1...Kxb6 2.b8Q/R+

1.a7? th. 2.a8G+
1...Kxb6 2.a8S+

1.h8nQ! zz
1...nQa8 2.nQxb7+
1...nQd8 2.nQc7+
1...nQf8 2.nQc5+
1...nQg8,nQg7 2.nQg2+
1...nQh6 2.nQc1+
1...nQh7 2.nQc2+
1...Gh~ 2.nQh1+
1...Gf~ 2.nQc3+
1...Ge7 2.nQe8+
1...bxa6 2.nQa8+
1...Kb5 2.nQxh5+
1...Kxb6 2.nQd8+
(1...nQb8 2.nQxb7,nQc7+)

Although the check after 1...Kxb6 is changed in three phases, the main weight is definitely given to the last phase, i.e. solution. Curiously, after nQ promotion there is no threat, but any legal move (with exception of 1...nQb8) leads to single possible legal check by White. Here the idea for composers might be in multiplying neutral promotions. It might work...

+2 (4+6+1)
1+4 grasshopper

Cornel Pacurar
Arno Tüngler

Comm 2010

13.Kxg1 27.Kxh3 30.Kg6 31.Rh5 32.Rg5 33.h5 34.Kh6 fxg5+

Captures of Sg1 and Ph3 unlock the knot so that bR can sacrifice for checking pawn capture.

ser-h+34 (6+10)

Bas de Haas
Probleemblad 2010

1.Kd6! zz 1...Kf3 2.Kd5+
1...Kh1 2.e4+
1...Kf1 2.d4+
1...Kh3 2.Ke5+
1...Kf2 2.NEg5+
1...Kh2 2.h5+
1...Kg1 2.NEd1+
1...Kg3 2.NEa3+

Similarly to the Karlström's twomover, bK has eight fligths and each of them is met by different checking move. What about changing eight checks?

+2 (8+2)
3+0 nonstop equihopper

Erich Bartel
dedicated to H. Gruber-50
Problemkiste 2010

1.a8Q a3 2.Qf8 a2 3.b8Q a1S 4.Qb1+ Sc2+

Kindergarden helpselfcheck is solved in a straightforward way: White promotes to two strongest pieces, queens, Black to the piece that can be forced to check. The question is whether the content can be multiplied in an interesting way.

hs+4 (3+2)

Ladislav Packa
Juraj Lörinc

2nd Prize R.J. Millour 70 JT 2014-15

1.Qe4? th. 2.Qb1 axb1S(Sg8)+
1...a1Q 2.Qxf4(Qd1) Qc1+
1...a1R 2.Qe1 Rxe1(Rh8)+
1...a1B 2.Qxd4(Qd1) Bxe5(Bf8)+
1...a1S 2.Qc2 Sxc2(Sg8)+

1.Sa8! th. 2.Qb1 axb1S(Sg8)+
1...a1Q 2.Kg5 Qxa8(Qd8)+
1...a1R 2.Qa7 Rxa7(Rh8)+
1...a1B 2.Qb2 Bxb2(Bf8)+
1...a1S 2.Qb3 Sxb3(Sg8)+

Here is an example of multiplying the content using powers offered by check as an aim. It is not necessary to care about all wK flights, only about allowing Black to check. How it works?

White queen wants to sacrifice on b1 for promotion capture. There are two possible ways there. In each case, wQ can sacrifice to R, B, S on different squares forcing the check. The key is the solving the Q promotion. In try wQ annihilates Pf4, in solution wS sacrifices itself on a8 in the key, just wK has to make one small move.

Z-24-48 with AUW defences is not a bad result. It is surely possible to work with other multiplication mechanisms.

r+2 (9+6)

Comments to Juraj Lörinc.
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