Final Award in Quick Composing TT-161 | Окончательные итоги блицконкурса TT-161

Asymmetrical solution | Асимметричное решение


Theme | Тема

51 entries were received from 14 authors representing 7 countries | На конкурс поступило 51 композиция от 14 авторов из 7 стран

EN <-> RU

51 entries were received for the tourney: 15 direct mates, 24 – helpmates, 12 – selfmates. This TT has a championship form in which participants compete in three sections. So, all non-excluded problems are scored with points. But in order to bring this championship to “usual” TT, I decided to give familiar distinctions additionally (Prizes, Hon. Mentions and Commendations) – it’s more clear and convenient for participants himself primarily.

I thank all participants for their creations and congratulate the Prize winners on a well-deserved success!

Overall ranking (evaluation of best #2-6 + evaluation of best h#2-6 + evaluation of best s#2-6 = overall evaluation):
1. Ingemar Lind (Sweden): 11+7.5+8.5=27
2-3. Sergey Abramenko (Russia): 12+9.5+5=26.5
2-3. Dieter Müller (Germany): 10.5+10.5+5.5=26.5
4. Ralf Krätschmer (Germany): 5+10+10=25
5. Pavel Arestov (Russia): 10+6+6.5=22.5
6. Rainer Kuhn (Germany): 7.5+6+6=19.5
7. József Molnár (Hungary): 6+6+5.5=17.5
8. Pietro Pitton (Italy): 8+9+0=17
9. Gábor Tar (Hungary): 5+11+0=16
10. Karol Mlynka (Slovakia): 0+8+5=13
11. Valery Liskovets (Belarus): 0+12+0=12
12. Michael Schlosser (Germany): 0+6.5+0=6.5
13-14. Christer Jonsson (Sweden): 0+6+0=6
13-14. Ladislav Packa (Slovakia): 0+6+0=6

Award is the following | Отличия распределились следующим образом

Direct mates | Задачи на прямой мат
15 entries were received from 10 authors representing 6 countries | На конкурс поступило 15 композиций от 10 авторов из 6 стран
EN <-> RU

15 problems competed in the #2-6 section. Only No 1 (Kf6-Kd6) was excluded in view of “non-thematic” symmetry.

12 points, 1st Prize, 1st Pl. - No 22
Sergey Abramenko
TT-161, SuperProblem, 23-03-2016
3K4/8/3B4/8/1P3P2/3Q4/3p4/3k4
#6(5+2)
11 points, 2nd Prize, 2nd Pl. - No 44
Ingemar Lind
TT-161, SuperProblem, 23-03-2016
4r3/4k3/4B3/4K3/8/8/4Q3/8
#3(3+2)
10.5 points, 3rd Prize, 3rd Pl. - No 8
Dieter Müller
TT-161, SuperProblem, 23-03-2016
8/4p3/4K3/4N3/4k3/4p3/4B3/4B3
#5**vvvv(4+3)

12 points, 1st Place, 1st Prize - No 22, Sergey Abramenko (Russia) 3K4/8/3B4/8/1P3P2/3Q4/3p4/3k4

1.f5? Kc1 2.Bf4 Kb2 3.Bxd2 – 4.Qc3+ Ka2 5.Bc1 Kb1 6.Qb2#, 1...Ke1! 2.Qe3 Kd1!
1.b5! zz 1...Ke1 2.Qe3+ Kf1 3.Qxd2 – 4.f5 Kg1 5.Bg3 Kh1 6.Qh2#, 3...Kg1 4.Qe2 Kh1 5.Qf1+ Kh2 6.f5#;
2...Kd1 3.Bb4 – 4.Qxd2#, 3...Kc2 4.Qc3+ Kb1 5.Qb3+ Ka1 6.Bc3#, 5...Kc1 Ba3#
It is very difficult choice of full value. It wants to play 1.f5? for the line-opening of wB to a “long” part of board. But only the line-opening of wB to a “short” part leads to a success. It is paradoxical and that is why it is interesting!
EN <-> RU

11 points, 2nd Place, 2nd Prize - No 44, Ingemar Lind (Sweden) 4r3/4k3/4B3/4K3/8/8/4Q3/8

1.Qf3? – 2.Qf6#, 1...R~/Rf8 2.Qf7/Qb7+ Kd8 3.Qd7#, 1...Kd8!
1.Qd3? – 2.Qd6#, 1...R~/Rd8 2.Qd7/Qh7+ Kf8 3.Qf7#, 1...Kf8!
1.Qd2? – 2.Qd6#, 1...Kf8 2.Qh6+ Ke7 3.Qf6#, 1...R~ 2.Qd7+ Kf8 3.Qf7#, 1...Rd8!
1.Qf2! – 2.Qf6#, 1...R~/Rf8 2.Qf7/Qa7+ Kd8 3.Qd7#, 1...Kd8 2.Qb6+ Ke7 3.Qd6#
In elementary position author manages to find a black correction with rook play. It is the only problem with a tactics – this fact explains a high place of the problem. In miniatures and gravures a short threat is not a factor which automatically reduces a value of the problem.
EN <-> RU

10.5 points, 3rd Place, 3rd Prize - No 8, Dieter Müller (Germany) 8/4p3/4K3/4N3/4k3/4p3/4B3/4B3

*1...Kd4 (a) 2.Sd7 (A) Ke4 3.Bg3 Kd4 4.Be5+ Ke4 5.Sc5#
*1...Kf4 (b) 2.Sf7 (B) Ke4 3.Bc3 Kf4 4.Be5+ Ke4 5.Sg5#
1.Bb4? (E) Kd4 (a) 2.Sf7 (B) Ke4 3.Bc3 Kf4 4.Be5+ Ke4 5.Sg5#, 1...Kf4! (b)
1.Bh4? (D) Kf4 (b) 2.Sd7 (A) Ke4 3.Bg3 Kd4 4.Be5+ Ke4 5.Sc5#, 1...Kd4! (a)
1.Ba5? (C) Kd4 (a) 2.Sd7 (A) Ke4 3.Bc7 Kd4 4.Be5+ Ke4 5.Sc5#, 1...Kf4! (b)
1.Sf7? (B) Kd4 (a) 2.Bb4 (E) Ke4 3.Bc3 Kf4 4.Be5+ Ke4 5.Sg5#, 1...Kf4! (b)
1.Sd7! (A)
1...Kd4 (a) 2.Ba5 Ke4 3.Bc7 Kd4 4.Be5+ Ke4 5.Sc5#
1...Kf4 (b) 2.Bh4 Ke4 3.Bg3 Kd4 4.Be5+ Ke4 5.Sc5#
The block-problem with elements of change play. It is good achievement for the TT theme. Rich choice of white play pleases.
EN <-> RU
10 points, Special Prize - No 26
Pavel Arestov
TT-161, SuperProblem, 23-03-2016
8/8/4Q3/4B3/4k3/8/4p3/4K3
#4(3+2)
9 points, Honorable mention - No 9
Dieter Müller
TT-161, SuperProblem, 23-03-2016
8/3p4/3B4/3B4/2PkP3/3P4/3R4/3K4
#5vvvvv(7+2)
8 points, Commendation - No 17
Pietro Pitton
TT-161, SuperProblem, 23-03-2016
3Q4/3K4/8/3k4/8/3p4/3N4/8
#4b) Qd8->d1(3+2)

10 points, Special Prize - No 26, Pavel Arestov (Russia) 8/8/4Q3/4B3/4k3/8/4p3/4K3

1.Kd2! Kf3 2.Qf5+ Kg2 3.Kxe2 Kg1/Kh1 4.Qf1#
1...e1Q+ 2.Kxe1 Kd3 3.Qg4! Ke3 4.Qe2#, 3…Kc2 4.Qd1#
1.Kf2? Kd3! 2.Qd5+ Kc2 3.Kxe2 Kb1! or 1.Kxe2? pat
Surely even Samuel Loyd would have praised the author for this finding! Chess amateurs (especially young) will certainly appreciate wit and subtlety of this 5-piece-problem. I always try to encourage interesting findings by “heightened” distinctions.
EN <-> RU

9 points, Honorable mention - No 9, Dieter Müller (Germany) 8/3p4/3B4/3B4/2PkP3/3P4/3R4/3K4

1.Kc2! – 2.Rf2 Ke3 3.Bc5#, 1...Ke3 2.Bf7 (A) Kf3 3.Bh5+ Ke3 4.Be5 d6/d5 5.Re2#
1.Ke2? – 2.Rb2 Kc3 3.Be5#, 1...Kc3!
1.Kc1? (B) Ke3!
1.Bf7? (A) Kc3!
1.Rf2? Kc3!
1.Re2? Kxd3 2.Bc5 Kc3 3.Kc1! (B) Kb3 4.Re3+ Ka2 5.Ra3#, 1...Kc3!
The merit of the problem is that White release the King of freedom via e3 where he has more “vast”, whereas it seems that it would better to do it via c3 – 1.Ke2? or 1.Re2?
EN <-> RU

8 points, Commendation - No 17, Pietro Pitton (Italy) 3Q4/3K4/8/3k4/8/3p4/3N4/8

a) diagram: 1.Qc7! Kd4 2.Ke6 Ke3 3.Qh2 Kd4 4.Qe5#
b) Qd8->d1: 1.Qg4! zz 1...Ke5 2.Ke7 Kd5 3.Qc4+ Ke5 4.Qe4#
1...Kc5 2.Kc7 Kd5 3.Qe4+ Kc5 4.Qc4#, 2...Kb5 3.Qc4+ Ka5 4.Sb3#
The twin that increases a scope of a play is appropriate. But there is a shortcoming – flight-taking keys, that is not a “fatal” shortcoming (it was noted in examples).
EN <-> RU
7.5 points, Special Comm. - No 41
Rainer Kuhn
TT-161, SuperProblem, 23-03-2016
4N3/3p1p2/3PkP2/3p1p2/1P2p2P/1Pp1K1pP/2P3P1/4R3
#6(11+8)
7 points - No 27
Pavel Arestov
TT-161, SuperProblem, 23-03-2016
8/3RpR2/4P3/4k3/8/8/8/4K3
#6(4+2)
7 points - No 45
Ingemar Lind
TT-161, SuperProblem, 23-03-2016
4k3/4p3/2p1K1p1/4p3/4Q3/3pbp2/4r3/8
#6(2+9)

7.5 points, Special Commendation - No 41, Rainer Kuhn (Germany) 4N3/3p1p2/3PkP2/3p1p2/1P2p2P/1Pp1K1pP/2P3P1/4R3

1.Kf4! d4 2.Ra1 Kd5 3.Ra6 (4.Sc7#) 3...d3 4.Sc7+ Kd4 5.Ra5 ~ 6.Rd5#
This problem has an idea – it is... schahographic, with Christmas tree!
EN <-> RU

7 points - No 27, Pavel Arestov (Russia) 8/3RpR2/4P3/4k3/8/8/8/4K3

1.Kd2! Ke4! 2.Ra7! Ke5 3.Ra6 Kd4 4.Rf5 Kc4 5.Rb6 Kd4 6.Rb4#, 2...Kd5 3.Ra6 Kc4 4.Rf5 Kb3 5.Rf4 Kb2 6.Rb4#;
1...Kxd6 2.Rdxe7+! Kd6 3.Kc3 Kc5 4.Rf6(Re6) Kb5 5.Ra7.
1.Kf2? Ke4!; 1.Ke2? Ke4!
It is very much unexpected key! If there were not small duals (that “erode” a design) then the problem would be stand higher.
EN <-> RU

7 points - No 45, Ingemar Lind (Sweden) 4k3/4p3/2p1K1p1/4p3/4Q3/3pbp2/4r3/8

1.Qxe5? – 2.Qb8/Qh8#, 1...B~!
1.Qxc6/Qxg6+? Kf8/Kd8!
1.Qh4? Kd8!
1.Qb4! Kf8 2.Qxe7+ Kg8 3.Qf7+ Kh8 4.Qf8+ Kh7 5.Kf7 – 6.Qg7#, 5...Bh6 6.Qg8#
Minimal problem. It is pleasant that author manages to find quiet moves for performing of the design.
EN <-> RU
6.5 points - No 21
Sergey Abramenko
TT-161, SuperProblem, 23-03-2016
1r2k2r/4p3/1p2P2p/1p5p/4R3/3P1P2/1p2K2p/4Q3
#5(6+10)
6.5 points - No 18
Pietro Pitton
TT-161, SuperProblem, 23-03-2016
8/8/8/8/8/1N1PkP1N/4P3/1R2K2R
#4(8+1)
6 points - No 34
József Molnár
TT-161, SuperProblem, 23-03-2016
4k3/2p3p1/4K3/2p3p1/4n3/8/8/4Q3
#2(2+6)

6.5 points - No 21, Sergey Abramenko (Russia) 1r2k2r/4p3/1p2P2p/1p5p/4R3/3P1P2/1p2K2p/4Q3

1.Qc3? 0-0!
1.Qg3! (2.Qb8#) 1...Rc8! 2.Rd4! Rc2+ 3.Ke3 Re2+ 4.Kxe2 – 5.Qb8#, 4...Kf8 5.Rd8#
White prevent a Black castling very straightforward.
EN <-> RU

6.5 points - No 18, Pietro Pitton (Italy) 8/8/8/8/8/1N1PkP1N/4P3/1R2K2R

1.0-0! Kxe2 2.Rbd1 Ke3 3.Kg2 Ke2 4.Rfe1#
And here, vice versa, White at once use their right for a castling.
EN <-> RU

6 points - No 34, József Molnár (Hungary) 4k3/2p3p1/4K3/2p3p1/4n3/8/8/4Q3

1.Qxe4? – 2.Qa8#, 1...c6!
1.Qh1! – 2.Qh8#, 1...Sf6 2.Qa8#
It is very simple but pleasant problem! I did not find direct anticipations in yacpdb and pdb.
EN <-> RU
5 points - No 5
Ralf Krätschmer
TT-161, SuperProblem, 23-03-2016
8/4Q3/4p3/3pkp2/4p3/4P3/4B3/4K3
#6(4+5)
5 points - No 38
Gábor Tar
TT-161, SuperProblem, 23-03-2016
8/4R3/8/8/4P3/3PkP2/4P3/4K3
#6(6+1)

5 points - No 5, Ralf Krätschmer (Germany) 8/4Q3/4p3/3pkp2/4p3/4P3/4B3/4K3

1.Qd8! f4 2.exf4+ Kxf4 3.Qf6+ Kg3 4.Qg5+ Kh3/Kh2 5.Bf1+ Kh2/Kh1 6.Qg2#

5 points - No 38, Gábor Tar (Hungary) 8/4R3/8/8/4P3/3PkP2/4P3/4K3

1.Rd7? Kf4!
1.Rf7? Kd4!
1.Re6! Kf4 2.Kf2 Kg5 3.Kg3 Kh5 4.Kf4 Kh4 5.Rh6#
1...Kd4 2.Kd2 Kc5 3.Kc3 Kb5 4.d4 Ka5 5.Kc4 Ka4 6.Ra6#, 4...Ka4 5.Re5 Ka3 6.Ra5#
Helpmates | Коопматы
24 helpmates were received from 14 authors representing 7 countries | На конкурс поступило 24 композиции от 14 авторов из 7 стран
EN <-> RU

I have received 24 problems in the h#2-6 section. Symmetrical location of pieces has an important nuance – “excess” vertical leftward of rightward. The ability to perform an asymmetrical solution bases on this fact. Here are little opportunities to realize the theme. Three types of situations can be distinguished:
- 1st type: bK moves on “short” edge of a board. It is the simplest. Such designs may be realized more easy and so its cannot be evaluated highly;
- 2nd type: Black or White piece perform a maneuver that is not possible in symmetrical try. This type is most interesting. Such compositions has maneuvers that emphasizes an asymmetry of a board and so its evaluated higher;
- 3rd type: castling (it is not permitted by conditions of the theme). The opportunity of White and/or Black castlings prompts a solution that is not very good.

Considering a specifics of the theme (asymmetrical solution in symmetrical position), the arbitrator evaluated the loading of White and Black pieces not strictly. The absence of tactic analogy in solution was not seen as a shortcoming at all, but the presence of this analogy was evaluated as a merit. At the same time, I used a strict approach to a way of twinning. The fact is that a presence of twins increases contents – it provides a competitive advantage. For example, shifted twins (with shifting a position to one vertical leftward of rightward) add a nothing to contents in view of a solution remains the same (mirror).

I have excluded three problems: No 14 (Ke7-Ke3) and No 51 (Ke4-Ke2) – because both solutions is symmetrical fully, and No 28 (Kd1-Kd7) in view of yacpdb/88365 (full anticipation).

12 points, 1st Prize, 1st Pl. - No 33
Valery Liskovets
TT-161, SuperProblem, 23-03-2016
8/4K3/4N3/3PbP1B/1pP3Pp/8/8/3k4
h#4(7+4)
11 points, 2nd Prize, 2nd Pl. - No 39
Gábor Tar
TT-161, SuperProblem, 23-03-2016
1r2k2r/2pp1pp1/2p3p1/4N3/3nPn2/8/4B3/4K3
h#32.1..(4+11)
10.5 points, 3rd Prize, 3rd Pl. - No 11
Dieter Müller
TT-161, SuperProblem, 23-03-2016
3n4/8/8/2pPp3/2PkP3/3N4/8/3K4
h#4b) –Sd3(5+4)

12 points, 1st Prize, 1st Place - No 33, Valery Liskovets (Belarus) 8/4K3/4N3/3PbP1B/1pP3Pp/8/8/3k4

1.Ke2 Be8 2.Kf3 g5 3.Kg4 Ba4 4.Kh5 Bd1#
Symmetry comes after 1st Black and White move (it is allowed by the conditions of the theme). After that bK moves on “short” edge of a board (1st type) and wB perform a maneuver with 2nd type. We see another two interesting effects: Platzwechsel and moving of pieces in rectangle d1-h5-e8-a4.
EN <-> RU

11 points, 2nd Prize, 2nd Place - No 39, Gábor Tar (Hungary) 1r2k2r/2pp1pp1/2p3p1/4N3/3nPn2/8/4B3/4K3

1.Rb5 Bxb5 2.Kd8 Sxc6+ 3.Kc8 Ba6#
1.0-0 Bh5 2.Kh8 Bxg6 3.Rg8 Sxf7#
All possible types are used: 1st type – in 1st solution, 2nd and 3rd types – in 2nd solution.
EN <-> RU

10.5 points, 3rd Prize, 3rd Place - No 11, Dieter Müller (Germany) 3n4/8/8/2pPp3/2PkP3/3N4/8/3K4

a) diagram: 1.Sc6 dxc6 2.Kxe4 c7 3.Kf3 c8Q 4.e4 Qh3#
b) –Sd3: 1.Se6 dxe6 2.Kc3 e7 3.Kb4 e8Q 4.Ka5 Qb5#
Twin is successful – presence and absence of Sd3 changes play cardinally. Model mates by Queens promoted on different squares.
EN <-> RU
10 points, 4th Prize - No 6
Ralf Krätschmer
TT-161, SuperProblem, 23-03-2016
2n1n3/p2b2p1/3k4/1r1B1r2/3p4/3K4/3R4/8
h#32.1..(3+9)
9.5 points, 5th Prize - No 24
Sergey Abramenko
TT-161, SuperProblem, 23-03-2016
3q4/3p4/3k4/1p1B1p2/1ppBpp2/1P3P2/8/3K4
h#32.1..(5+9)
9 points, Special Prize - No 20
Pietro Pitton
TT-161, SuperProblem, 23-03-2016
1r2k2r/8/8/4K3/8/8/4P3/8
h#5(2+3)

10 points, 4th Prize - No 6, Ralf Krätschmer (Germany) 2n1n3/p2b2p1/3k4/1r1B1r2/3p4/3K4/3R4/8

1.Kc7 Rc2+ 2.Kb8 Rc6 3.Ka8 Rxc8#
1.Sc7 Bh1 2.Rfc5 Rg2 3.Kc6 Rg6#
There is an analogy between solutions (creation of battery). Two types of asymmetry are realized.
EN <-> RU

9.5 points, 5th Prize - No 24, Sergey Abramenko (Russia) 3q4/3p4/3k4/1p1B1p2/1ppBpp2/1P3P2/8/3K4

1.exf3 Bxf3 2.Ke7 Bc5+ 3.Ke8 Bh5#
1.Kc7 Bf6 2.Kb6 Bb7 3.Ka5 Bxd8#
“Worker” problem! As in previous problem, here two types of asymmetry are realized.
EN <-> RU

9 points, Special Prize - No 20, Pietro Pitton (Italy) 1r2k2r/8/8/4K3/8/8/4P3/8

1.0-0 e4 2.Rf5+ exf5 3.Rb7 f6 4.Rh7 f7+ 5.Kh8 f8Q#
Excellent 5-piece-problem with excelsior and final model mate!
EN <-> RU
8 points, 1st Hon. mention - No 3
Karol Mlynka
TT-161, SuperProblem, 23-03-2016
2nKn3/2PPP3/2PrP3/3P4/3k4/8/8/8
h#32.1..(7+4)
7.5 points, 2nd Hon. mention - No 19
Pietro Pitton
TT-161, SuperProblem, 23-03-2016
3r1r2/8/4k3/8/8/8/1PP3PP/1R2K2R
h#3b) Ke6->e7(7+3)
7.5 points, 3rd Hon. mention - No 47
Ingemar Lind
TT-161, SuperProblem, 23-03-2016
4k3/3p1p2/8/4p3/1p2N2p/4K3/4R3/4b3
h#32.1..(3+7)

8 points, 1st Honorable mention - No 3, Karol Mlynka (Slovakia) 2nKn3/2PPP3/2PrP3/3P4/3k4/8/8/8

1.Rxc6 dxe8Q 2.Ke5 Qh5+ 3.Kf6 e8S#
1.Kc5 Kxe8 2.Kb6 dxc8Q 3.Ka7 Qb7#
There is some connection between solutions (promotions), but 1st solution is more unexpected. In the problem there is a large choice of moves that it certainly increases its quality.
EN <-> RU

7.5 points, 2nd Honorable mention - No 19, Pietro Pitton (Italy) 3r1r2/8/4k3/8/8/8/1PP3PP/1R2K2R

a) diagram: 1.Rd6 c4 2.Rfd8 0-0 3.R8d7 Rbe1#
b) Ke6->e7: 1.Rfe8 0-0 2.Rc8 Rf7+ 3.Kd8 Rd1#
Two dissimilar mates are good. White use the same resource (a castling) on different moves – it is not good.
EN <-> RU

7.5 points, 3rd Honorable mention - No 47, Ingemar Lind (Sweden) 4k3/3p1p2/8/4p3/1p2N2p/4K3/4R3/4b3

1.b3 Rf2 2.Ba5 Rxf7 3.Bd8 Sd6#
1.Kf8 Sf6 2.Kg7 Rg2+ 3.Kh8 Rg8#
Symmetrical plans do not lead to success - 1.b3 Rd2 2.Bb4?? and 1.h3 Rf2 2.Bh4?? Besides that, after 1.h3 Rd2 bB has not access to square similar a5.
EN <-> RU
7 points, Special HM - No 32
Valery Liskovets
TT-161, SuperProblem, 23-03-2016
4K3/4B3/1r2B2r/3PpP2/2P3P1/3P1P2/4q3/4k3
h#4b) –Be7, –Be6(9+5)
7 points, Special HM - No 46
Ingemar Lind
TT-161, SuperProblem, 23-03-2016
1r1nkn1r/4b3/3p1p2/3PNP2/8/2P3P1/8/1R2K2R
h#2b) Ke8->e5(8+8)
7 points, Special HM - No 2
Karol Mlynka
TT-161, SuperProblem, 23-03-2016
3K4/8/3R4/3b4/8/8/8/3k4
h#4b) a1=b1(2+2)

7 points, Special Honorable mention - No 32, Valery Liskovets (Belarus) 4K3/4B3/1r2B2r/3PpP2/2P3P1/3P1P2/4q3/4k3

a) diagram: 1.Kf2 Bd7 2.Kxf3 Ba4! 3.Kxg4 Bd1 4.Kh5! Bxe2#
b) –Be7, –Be6: 1.Kd2 f6 2.Kc3 f7 3.Kb4 f8Q+ 4.Ka5! Qa3#!
This and next problem claim to have a large-scale content. But a way of twinning is bad. Especially it is evident here – wBe6 is removed from a board only for model mate achievement.
EN <-> RU

7 points, Special Honorable mention - No 46, Ingemar Lind (Sweden) 1r1nkn1r/4b3/3p1p2/3PNP2/8/2P3P1/8/1R2K2R

a) diagram: 1.Sd7 Rb4 2.0-0 Rg4#
b) Ke8->e5: 1.Rb5 0-0 2.Rxd5 Rbe1#

7 points, Special Honorable mention - No 2, Karol Mlynka (Slovakia) 3K4/8/3R4/3b4/8/8/8/3k4

a) diagram: 1.Kc2 Kc7 2.Kb3 Kb6 3.Ka4 Kc5 4.Bb3 Ra6#
b) a1=b1: 1.Kf2 Kf7 2.Kg3 Kg6 3.Kh4 Kf5 4.Bg3 Rh6#
It is the only “survivor” 4-piece-problem in the section! The evaluation is given without regard to a twin.
EN <-> RU
6.5 points, Commendation - No 10
Dieter Müller
TT-161, SuperProblem, 23-03-2016
3K4/3b4/1p1p1p2/3k4/3B4/8/8/3B4
h#2.52.1..(3+5)
6.5 points, Commendation - No 23
Sergey Abramenko
TT-161, SuperProblem, 23-03-2016
1r2k2r/8/8/4B3/2P1b1P1/4N3/8/4K3
h#2b) wRe3(5+4)
6.5 points, Commendation - No 50
Michael Schlosser
TT-161, SuperProblem, 23-03-2016
8/8/2p1k1p1/8/8/4B3/2p1R1p1/4K3
h#32.1..(3+5)

6.5 points, Commendation - No 10, Dieter Müller (Germany) 3K4/3b4/1p1p1p2/3k4/3B4/8/8/3B4

1...Bh5 2.Bf5 Ke8 3.Ke6 Bf7#
1...Bc2 2.Ke6 Bh7 3.f5 Bg8#

6.5 points, Commendation - No 23, Sergey Abramenko (Russia) 1r2k2r/8/8/4B3/2P1b1P1/4N3/8/4K3

a) diagram: 1.Bh7 Sf5 2.0-0 Sh6#
b) wRe3: 1.0-0 Rh3 2.Rf7 Rh8#

6.5 points, Commendation - No 50, Michael Schlosser (Germany) 8/8/2p1k1p1/8/8/4B3/2p1R1p1/4K3

1.Kf5 Bf2 2.Kg4 Re4+ 3.Kh3 Rh4#
1.Kf7 Bd4 2.Kg8 Re8+ 3.Kh7 Rh8#
6 points, Commendation - No 15
Christer Jonsson
TT-161, SuperProblem, 23-03-2016
8/3N4/p2P2p1/8/3k4/3n4/3n4/3K4
h#4b) Pa6->a3, Pg6->g3
c) Sd2->c5, Sd3->e5
d) Pa6->b7, Pg6->f7
(3+5)
6 points - No 16
Ladislav Packa
TT-161, SuperProblem, 23-03-2016
4n3/2ppppp1/4k3/2r3r1/2P3P1/2pP1Pp1/8/1R2K2R
h#2.52.1..(7+11)
6 points - No 29
Pavel Arestov
TT-161, SuperProblem, 23-03-2016
8/2prp3/8/3k4/3P4/8/3p4/3K4
h#5(2+5)

6 points, Commendation - No 15, Christer Jonsson (Sweden) 8/3N4/p2P2p1/8/3k4/3n4/3n4/3K4

a) diagram: 1.Kd5 Sb6+ 2.Kc6 d7 3.Kb7 d8Q 4.Ka7 Qc7#
b) Pa6->a3, Pg6->g3: 1.Kc3 Sc5 2.Kb2 Kxd2 3.Ka1 Kc2 4.a2 Sb3#
c) Sd2->c5, Sd3->e5: 1.Sb7 Sxe5 2.Kc5 d7 3.Kb6 d8Q+ 4.Ka7 Sc6#
d) Pa6->b7, Pg6->f7: 1.Kd5 Sf6+ 2.Ke6 d7 3.Ke7 Sh5 4.Kf8 d8Q#
Elaboration of earlier problem of the author – yacpdb/404118. This problem claims to have a large-scale content. But a way of twinning is bad.
EN <-> RU

6 points - No 16, Ladislav Packa (Slovakia) 4n3/2ppppp1/4k3/2r3r1/2P3P1/2pP1Pp1/8/1R2K2R

1...0-0 2.Ke5 Rbe1+ 3.Kf4 Re4#
1...Rh2 2.Kd6 Ra2 3.Kc6 Ra6#

6 points - No 29, Pavel Arestov (Russia) 8/2prp3/8/3k4/3P4/8/3p4/3K4

1.Kc6 d5+ 2.Kb7 d6 3.Rd8 dxc7 4.Rb8 c8Q+ 5.Ka8 Qa6#
6 points - No 35
József Molnár
TT-161, SuperProblem, 23-03-2016
3K4/1p1b1p2/1P1B1P2/3p4/3k4/1P1p1P2/1P3P2/8
h#3(8+6)
6 points - No 42
Rainer Kuhn
TT-161, SuperProblem, 23-03-2016
3k4/8/8/8/3n4/3p4/3Q4/3K4
h#4b) –Sd4, +Rd6(2+3)
5 points - No 36
József Molnár
TT-161, SuperProblem, 23-03-2016
8/8/8/3K4/3B4/2N1N3/3k4/8
h#31.2.1..(4+1)

6 points - No 35, József Molnár (Hungary) 3K4/1p1b1p2/1P1B1P2/3p4/3k4/1P1p1P2/1P3P2/8

1.Bg4 Bh2 2.Bxf3 Bg1 3.Be4 f4#

6 points - No 42, Rainer Kuhn (Germany) 3k4/8/8/8/3n4/3p4/3Q4/3K4

a) diagram: 1.Kc7! Qxd3 2.Kb6 Kd2 3.Ka5 Kc3 4.Ka4 Qa6#
b) –Sd4, +Rd6: 1.Kc7! Qf4 2.Kb6 Kd2 3.Ka5 Kc3 4.Ra6 Qb4#

5 points - No 36, József Molnár (Hungary) 8/8/8/3K4/3B4/2N1N3/3k4/8

1.Kc1 Bc5 2.Kb2 Ba3+ 3.Ka1 Sc2#
1...Kc4 2.Kb2 Sed1+ 3.Ka3 Bc5#
Selfmates | Самоматы
12 selfmates were received from 9 authors representing 5 countries | На конкурс поступило 12 композиции от 9 авторов из 5 стран
EN <-> RU

12 problems competed in the s#2-6 section. I have excluded No 13 (Ke1-Ke3) in view of yacpdb/403833 and No 40 (Ke1-Ke3) in view of symmetrical solution. The rest problems are evaluated with points.

10 points, Prize, 1st Place - No 7
Ralf Krätschmer
TT-161, SuperProblem, 23-03-2016
8/4Q3/2N3N1/4p3/4r3/4k3/4P3/3RKR2
s#6(7+3)
8.5 points, HM, 2nd Place - No 49
Ingemar Lind
TT-161, SuperProblem, 23-03-2016
8/3p4/3R4/NPPBPPN1/1PPpPP2/3k4/3B4/3K4
s#5(14+3)
7 points, Comm., 3rd Place - No 48
Ingemar Lind
TT-161, SuperProblem, 23-03-2016
8/8/1pP3Pp/8/1P1pBp1P/1P1PkP1P/4p3/1N2K2N
s#5(12+6)

10 points, Prize, 1st Place - No 7, Ralf Krätschmer (Germany) 8/4Q3/2N3N1/4p3/4r3/4k3/4P3/3RKR2

1.Rd5! zz 1...R~ 2.Rxe5+ Re4 3.Rf4 zz Rxe5 4.Qa7+ Rc5 5.Qa3+ Rc3 6.Qc1+ Rxc1#
The choice of 1st move became clear only after 4.Qa7+! If 1.Rf5? then wQ has not the same long move (as in solution).
EN <-> RU

8.5 points, Hon. mention, 2nd Place - No 49, Ingemar Lind (Sweden) 8/3p4/3R4/NPPBPPN1/1PPpPP2/3k4/3B4/3K4

1.Bc6! dxc6 2.Rh6 cxb5 3.Rh1 bxc4 4.Re1 c3 5.Bc1 c2#
Black pawn starts from initial to “prize” square. All play is quiet.
EN <-> RU

7 points, Commendation, 3rd Place - No 48, Ingemar Lind (Sweden) 8/8/1pP3Pp/8/1P1pBp1P/1P1PkP1P/4p3/1N2K2N

1.h5! b5 2.Bd5 Kxd3 3.Be4+ Ke3 4.Sa3 d3 5.Sxb5 d2#
Although temporarily, bK finds a freedom.
EN <-> RU
6.5 points - No 30
Pavel Arestov
TT-161, SuperProblem, 23-03-2016
4Q3/3R1R2/8/4p3/4k3/8/4B3/4K3
s#4(5+2)
6 points - No 31
Pavel Arestov
TT-161, SuperProblem, 23-03-2016
4Q3/8/4p3/8/1R5R/3PkP2/4B3/4K3
s#4(7+2)
6 points - No 43
Rainer Kuhn
TT-161, SuperProblem, 23-03-2016
8/4Q3/1p2N2p/1P2p2P/2P1q1P1/4k3/2P1p1P1/4K3
s#4(9+6)

6.5 points - No 30, Pavel Arestov (Russia) 4Q3/3R1R2/8/4p3/4k3/8/4B3/4K3

1.Rd1! Ke3 2.Qa4 e4 3.Rf3+ exf3 4.Bf1 f2#

6 points - No 31, Pavel Arestov (Russia) 4Q3/8/4p3/8/1R5R/3PkP2/4B3/4K3

1.Qa4! e5 2.Qd1 e4 3.d4 exf3 4.Bf1 f2#

6 points - No 43, Rainer Kuhn (Germany) 8/4Q3/1p2N2p/1P2p2P/2P1q1P1/4k3/2P1p1P1/4K3

1.Qa3+! Qd3 2.cxd3 e4 3.Qd6 exd3 4.Qc6 d2#
5.5 points - No 12
Dieter Müller
TT-161, SuperProblem, 23-03-2016
3Q4/8/3N4/1R3R2/3k4/8/3K4/1r3r2
s#4b) a1=b1(5+3)
5.5 points - No 37
József Molnár
TT-161, SuperProblem, 23-03-2016
3q4/3B4/2R1R3/3k4/3P4/2PKP3/3P4/3Q4
s#4(9+2)
5 points - No 4
Karol Mlynka
TT-161, SuperProblem, 23-03-2016
8/2p1B1p1/2P1p1P1/8/3ppp2/4k3/2P1p1P1/4K3
s#3(6+8)

5.5 points - No 12, Dieter Müller (Germany) 3Q4/8/3N4/1R3R2/3k4/8/3K4/1r3r2

a) diagram: 1.Rf4+! Rxf4 2.Qh8+ Rf6 3.Qh4+ Rf4 4.Qf2+ Rxf2#
b) a1=b1: 1.Rc4+! Rxc4 2.Qa8+ Rc6 3.Qa4+ Rc4 4.Qc2+ Rxc2#
Reciprocal change of bR’s functions (guard/mate) and wR’s (sacrifice/guard). The evaluation is given without regard to a twin.
EN <-> RU

5.5 points - No 37, József Molnár (Hungary) 3q4/3B4/2R1R3/3k4/3P4/2PKP3/3P4/3Q4

1.Qh5+ Qg5 2.Re5+ Qxe5 3.Qf5 zz Qxf5+ 4.e4+ Qxe4#

5 points - No 4, Karol Mlynka (Slovakia) 8/2p1B1p1/2P1p1P1/8/3ppp2/4k3/2P1p1P1/4K3

*1...e5 2.Bc5 (C) zz f3 3.g3 f2# / 2.Bg5 (D) zz d3 3.c3 d2#
1.Ba3? zz e5 2.Bc5 (C) zz 3.g3 f2#, 1...f3! / 1...d3!
1.Bh4? zz e5 2.Bg5 (D) zz d3 3.c3 d2#, 1...f3! / 1...d3!
1.Bc5! (C) zz e5 2.Ba7 (A) zz f3 3.g3 (B) f2#, 1...f3 2.g3 (B) zz e5 3.Ba7 (A) f2#
5 points - No 25
Sergey Abramenko
TT-161, SuperProblem, 23-03-2016
3RBR2/8/2p3p1/4B3/2P1p1P1/4k3/8/4K3
s#5(7+4)

5 points - No 25, Sergey Abramenko (Russia) 3RBR2/8/2p3p1/4B3/2P1p1P1/4k3/8/4K3

1.Rf1! c5 2.Ba4 (A) g5 3.Rd4 (B) cxd4 4.Bc2 d3 5.Bd1 d2#
1...g5 2.Rd4 (B) c5 3.Ba4 (A) cxd4 4.Bc2 d3 5.Bd1 d2#

URL address of this web page | Адрес этой страницы http://superproblem.ru/htm/tourneys/quick-tt/results/tt-161_award.html




Sections | Разделы

#2-6 | h#2-6 | s#2-6

Participants | Участники

Abramenko S. – No 21 (#5), 22 (#6), 23 (h#2), 24 (h#3), 25 (s#5)
Arestov P. – No 26 (#4), 27 (#6), 28 (h#4), 29 (h#5), 30 (s#4), 31 (s#4)
Jonsson C. – No 14 (h#3.5), 15 (h#4)
Krätschmer R. – No 5 (#6), 6 (h#3), 7 (s#6)
Kuhn R. – No 41 (#6), 42 (h#4), 43 (s#4)
Lind I. – No 44 (#3), 45 (#6), 46 (h#2), 47 (h#3), 48 (s#5), 49 (s#5)
Liskovets V. – No 32 (h#4), 33 (h#4)
Mlynka K. – No 1 (#3), 2 (h#4), 3 (h#3), 4 (s#3)
Molnár J. – No 34 (#2), 35 (h#3), 36 (h#3), 37 (s#4)
Müller D. – No 8 (#5), 9 (#5), 10 (h#2.5), 11 (h#4), 12 (s#4), 13 (s#5)
Packa L. – No 16 (h#2.5)
Pitton P. – No 17 (#4), 18 (#4), 19 (h#3), 20 (h#5)
Schlosser M. – No 50 (h#3), 51 (h#4)
Tar G. – No 38 (#6), 39 (h#3), 40 (s#4)

The Winners Are | Победители

Sergey Abramenko (#2-6)

Valery Liskovets (h#2-6)

Ralf Krätschmer (s#2-6)

Ingemar Lind
(overall ranking | общий зачет)

Congratulations! | Поздравляем!

Judge | Арбитр

Igor Agapov

Director and editor
Директор и редактор

Aleksey Oganesjan
e-mail: alexeioganesyan@gmail.com