Final Award in Quick Composing TT-240 | Окончательные итоги блицконкурса TT-240

Pata morgana | Pata morgana

Theme | Тема

154 entries were received from 33 authors representing 22 countries | На конкурс поступило 154 композиций от 33 авторов из 22 стран

EN <-> RU

After careful learning of all received problems of this TT I had a few thoughts about its theme, and I want to share these thoughts:
1) preferably a beginning side has more than one move to choose in diagram position. Otherwise the theme loses its paradox and also – in some cases – you may feel that the problem is artificially “tailored” to the theme. This does not mean that I didn’t consider the problems with the only first move as pretenders for an award. But in order to be included in the award, the first move must be an integral part of the concept;
2) and also about pointed artificiality. There were some problems with good content but one or several initial moves were clearly out of the concept. For example, No 101 (Kh1-Kb8) is h=6.5, where initial 2.5 moves are repetitive in both solutions and initial White move are the only. The way out is obvious: the author should remove these initial 2.5 moves and send a resulting h=4 for some another tourney where the problem can get a distinction in view of it contains total mixed AUW;
3) the problem with the set theme has an usual and unique property – both in initial and final positions the same side is stalemated. That’s why several authors at once guessed to realize the theme in twins form where a final position of one twin is an initial position of following! In view of complexity, I included all such problems in the award (there were three);
4) there were sent many problems in which a beginning side by some simple known maneuver with losing of tempo (for example, any kinds of triangle method) passes the move turn to an opponent in order to set the same (or very similar) stalemate as in diagram position. If such problems had no any additional interesting moments then I didn’t consider its;

A presence of interesting idea, concept (even maybe small but clear and interesting) was a main criteria for including a problem in the award. Otherwise you may feel that the author accidentally placed some stalemate position and (by changing a stipulation) searched the position that would has the only solution.
That is why the award is very strong – from 154 problems I selected only 24. In view of how these 24 problems “distributed” by genres I decided to do a common award without separating by genres.

After notes for a provisional award, the problem No 138 (Kc6-Kc8), initially marked by Special Comm., is excluded in view of full anticipation – pdb/P1328366.

Award is the following | Отличия распределились следующим образом

1st Prize - No 109
Ivo Tominić
TT-240, SuperProblem, 09-07-2020
hs=3.52 sol.(3+10)
2nd Prize - No 2
Boris Shorokhov
TT-240, SuperProblem, 09-07-2020
h=2.54 sol.(12+10)
3rd-4th Prize - No 63
Jorma Paavilainen
TT-240, SuperProblem, 09-07-2020
h=2.52 sol.(11+9)

1st Prize - No 109, Ivo Tominić (Croatia) 8/4b3/r7/ppRp1k2/3qp3/K2B1r2/8/2n5

1...Rf4 2.Bc4 Rg6 3.Bxd5 Qf6 4.Be6+ Kxe6=
1...Bg5 2.Rc4 Qe5 3.Rxe4 Re6 4.Rf4+ Kxf4=
In diagram position both White pieces are pinned. In each solution three of four (!) Black thematic pieces perform square-blockings around bK and also first of these pieces unpins White piece on the way. This unpinned piece become a front piece of formed battery and execute a “shot” by double check in order to force bK to capture it with stalemate because rear battery piece remains pinned. Excellent problem with good economy and loading of pieces of both sides and also with its clear functions permutation. I congratulate the author with a creative luck!
EN <-> RU

2nd Prize - No 2, Boris Shorokhov (Russia) 8/2p5/p1P5/p1PNN3/B1P3p1/1nP2pP1/Rnk1pP2/R2bK3

1...Ra3 2.Sxc4 (S2~?) R3a2+ 3.Sb2 c4=
1...Bb5 2.Sxc5
(S3~?) Ba4+ 3.Sb3 c5=
(Sd~?) 2.Kxc3 Sd5+ 3.Kc3-c2 c7=
(Se~?) 2.Kd3 Se5+ 3.Kd3-c2 g4=
When I composed a schematic example for the announcement of this TT, I want to get a problem with four (or at least three) thematic White pieces annihilating Black Pawns for providing tempo-moves. I didn't manage it offhand, unlike an author of this problem which chose a little the other way: in two solutions an annihilation is executed by White pieces and in two another – by Black! As a result we see a very clear HOTF with switchbacks of pieces of both sides.
EN <-> RU

3rd-4th Prize - No 63, Jorma Paavilainen (Finland) 8/4p3/p2pP3/P1pP2p1/R1b2kN1/Pp4NK/1P1n2P1/2B5

1...Se4 2.Bd3 Sg3+ 3.Be4 Rc4=
1...Se3 2.Sf1 Sg4+ 3.Se3 Bd2=
In diagram position both Black pieces are pinned. In each solution White piece indirectly unpins Black, after that is returns and as a result pinning and pinned pieces “shifts” on another squares of the pin line. There is a shortcoming – in view of bPc5, final White move has heterogeneous motif: Rc4 – blocking of this Pawn, Bd2 – tempo.
EN <-> RU
3rd-4th Prize - No 93
Henry Tanner
TT-240, SuperProblem, 09-07-2020
h=2.52 sol.(9+11)
5th Prize - No 111
Menachem Witztum & Emanuel Navon
TT-240, SuperProblem, 09-07-2020
hs=2.5b) wRe6(4+15)
6th Prize - No 105
János Csák
TT-240, SuperProblem, 09-07-2020
hs=3.5b) Kf4->e3(4+12)

3rd-4th Prize - No 93, Henry Tanner (Finland) 8/8/4p1p1/4PbP1/2N1pBpK/2N1npP1/2p2p1P/2k2n1R

1...Sd2 2.Sd1 Sc4+ 3.Sfe3 Rf1=
1...Sd1 2.Sd2 Sc3+ 3.Sef1 Be3=
Sharing of distinction for this and previous problems is quite explainable – schemes are quite similar. But in this problems the motifs are somewhat different: indirectly unpinned first Black piece, in its turn, itself indirectly unpins a second piece which after that moves on the place of first the one. Small shortcomings are (comparing to previous problem): wK doesn’t participate in the stalemate and bBf5 (for guarding g4) is need only in one solution.
EN <-> RU

5th Prize - No 111, Menachem Witztum & Emanuel Navon (Israel) 1q4n1/brpK4/pkpNN3/p1p5/P3p1b1/2p2p2/3r4/8

a) diagram: 1...Bh5 2.Sd4 Bf7 3.Sc4+ Bxc4=
b) wRe6: 1...Rh2 2.Sf5 Rh6 3.Rxc6+ Rxc6=
The beginning of this problem is similar to No 93, but the final is completely different from both No 93 and 63 – here we see a sacrifice of second White piece. On the whole, all is made quite clearly and with elegant twin but all this is somewhat simpler that in two pointed problems.
EN <-> RU

6th Prize - No 105, János Csák (Hungary) 7b/4p3/4n3/8/5kpp/p6p/PN1n1p1p/K1N4r

a) diagram: 1...Sd4 2.Sbd3+ (A) Kg3 3.Se1 e5 4.Se2+ (B) Sxe2=
b) Kf4->e3: 1...f1B 2.Se2 (B) Kf2 3.Sc3 Ke1 4.Sbd3+ (A) Bxd3=
And also we see the same motifs of “relay race” indirect unpinning that executed by different sides. Here it is supplemented by reciprocal change of moves. The problem could be evaluated higher but I was very confused by absence of full-value permutation of functions between Se6 and Pf2 – in b) this Knight is unnecessary at all.
EN <-> RU
7th Prize - No 108
Henry Tanner
TT-240, SuperProblem, 09-07-2020
b) final pos. of a)
Special Prize - No 120
János Csák
TT-240, SuperProblem, 09-07-2020
a)Ph7->f7; b)Pf6->h6
1st Honorable mention - No 114
Marko Klasinc
TT-240, SuperProblem, 09-07-2020

7th Prize - No 108, Henry Tanner (Finland) 6k1/3b2P1/3p2PB/3P2P1/4K1p1/5p2/3q4/8

a) diagram: 1...Bc6 2.dxc6 f2 3.c7 f1B 4.c8B Bc4 5.Be6+ Bxe6=
b) final pos. of a): 1...Bf7 2.gxf7+ Kxf7 3.g6+ Kxg6 4.g8Q+ Kxh6 5.Qg6+ Kxg6=
It is the best of three sent problems with unique type of twins – final position of one twin is initial position of following. The author managed to achieve a good economy and loading of pieces and also a worthy homogeneity in the play: each phase begins with a sacrifice of bB under wP and ends with a sacrifice of White promoted piece.
EN <-> RU

Special Prize - No 120, János Csák (Hungary) k7/1pP4p/1P3P2/8/8/1p6/1p6/bK6

a) Ph7->f7: 1.c8S! Kb8 2.Sd6 Ka8 3.Sxf7 Kb8 4.Sd6 Ka8 5.f7 Kb8 6.f8S! Ka8 7.Sd7=
b) Pf6->h6: 1.c8B! Kb8 2.Bf5 Ka8 3.Bxh7 Kb8 4.Bf5 Ka8 5.h7 Kb8 6.h8B! Ka8 7.Be5=
Solution of this problem is quite nice and clear without additional explanations. I am quite loyal to zero form if it is justified by complex and/or record content, and here we see such case – gravure with two underpromotions in the same square in each phase. But zero is zero, and that’s why I can get only special distinction for this problem.
EN <-> RU

1st Honorable mention - No 114, Marko Klasinc (Slovenia) 8/5b2/6p1/5kP1/6pB/5pKR/5P1P/8

1...Ke6! 2.Kf4 Be8! 3.Rg3 Bb5 4.Rg1 Bf1! 5.Kg3 Bh3 6.Re1+ Kf5 7.Re5+ Kxe5 =
The problem with releasing a square “a la Klasinc” and very interesting motivation of maneuvers: ultimately bB must stand on h3, that’s why wR must get out of there, and a square g3 for it must be released by wK, and for this bK must let it out through f4. And since the only way to do it – 1…Ke6, this move is anti-ziel element that forced bB to spend one move more during a route to h3. In the future the move 4…Bf1 also can be formally named as anti-ziel (because wR is need to e1) but, since before a move Re1 White is need to provide bK a move to f5, a “respite” for wR turns out to be quite handy. It is very curious “plot” problem with good geometric scope in the play.
EN <-> RU
2nd Honorable mention - No 125
Alberto Armeni
TT-240, SuperProblem, 09-07-2020
3rd Honorable mention - No 103A
Karol Mlynka
TT-240, SuperProblem, 09-07-2020
hs=3.53 sol.(4+7)
4th-5th Honorable mention - No 48
Menachem Witztum
TT-240, SuperProblem, 09-07-2020
h=4.52 sol.(9+4)

2nd Honorable mention - No 125, Alberto Armeni (Italy) 8/8/1p1Q4/1P6/4k1P1/5p2/5K2/8

1.Qd1? Kf4 2.Qd5 Kxg4 3.Qe5 Kh4 4.Qf5=, 3...Kh3 4.Qf4=, 1...Ke5!
1.g5! Kf5 2.Kxf3 Kxg5 3.Qe6 Kh5 4.Qf6=, 3...Kh4 4.Qf5=
Very clear and nice echo-play in try and solution. Additionally in both phases there is a flight-giving-key.
EN <-> RU

3rd Honorable mention - No 103A, Karol Mlynka (Slovakia) 2n1k2K/3bPq2/2p1p3/2P3b1/6P1/8/8/8

1...Bh6 2.g5 Bf8 3.g6 Qxg6 4.exf8Q+ Ke8xf8=
1...Qf6+ 2.Kh7 Kf7 3.e8R Sa7 4.Rf8+ Kf7xf8=
1...Sd6 2.cxd6 Bc8 3.d7+ Kxe7 4.d8Q+ Ke7xd8=
Three solutions are united by FML effects, White promotions with sacrifices and final acceptances of these sacrifices by bK.
EN <-> RU

4th-5th Honorable mention - No 48, Menachem Witztum (Israel) 7k/1p2N1p1/1P4P1/KP6/PPp5/2P5/2P5/8

1...Sc6 2.bxc6 b7 3.c5 b8S 4.cxb4 Sc6 5.bxc3 Se7=
1...Sf5 2.Kg8 Sxg7 3.Kf8 Sf5 4.Ke8 g7 5.Kd7 g8Q=
Solutions are united by switchback of White Knight (initial or promoted) and White promotions. bK walk out from the corner is quite unexpected.
EN <-> RU
4th-5th Honorable mention - No 96
Henry Tanner
TT-240, SuperProblem, 09-07-2020
b) final pos. of a)
6th Honorable mention - No 117
Marko Klasinc
TT-240, SuperProblem, 09-07-2020
b,c) final pos. of a,b)
Special Honorable mention - No 131
Solaiappan Manikumar
TT-240, SuperProblem, 09-07-2020

4th-5th Honorable mention - No 96, Henry Tanner (Finland) 4k3/2p1P1p1/2P3Pp/5N1P/6p1/6p1/6Pp/7K

a) diagram: 1...Sxh6 2.Kxe7 Sf5+ 3.Kf8 h6 4.Kg8 hxg7=
b) final pos. of a): 1...Sd6 2.Kxg7 Se4 3.Kf8 g7+ 4.Ke7 g8Q=
The scheme and motifs are very similar to previous problem.
EN <-> RU

6th Honorable mention - No 117, Marko Klasinc (Slovenia) 3k1KBb/3P1P1P/7p/4n2p/5pnp/8/7p/8

a) diagram: 1...Bf6 2.h8R Bg5 3.Kg7 Ke7 4.Kh7 Kf8 5.d8Q+ Bxd8 =
b) final pos. of a): 1...Sf6+ 2.Kxh6 Sxg8+ 3.Kxh5 Kg7 4.f8Q+ Kxh8 5.Qg7+ Kxg7 =
c) final pos. of b): 1...Sf3 2.Kg4 Bb6 3.Kh3 Bg1 4.Kg2 h3+ 5.Kh1 ~ =
It is worth noting that the overall uniformity of the play in this problem is not very good: yes, a) and b) are united (as in No 112 of the same author) by sacrifices of White promoted pieces in the final, but in the final of c) there is no any similar. Instead it we see a stalemate by random Black move. Is this permissible in hs# and hs=? It is well-known matter of dispute. Personally, I think that is unacceptable, because these genres is very close to h# and h= where such duals on mating (stalemating) move are forbidden traditionally. But it is only my personal point of view.
However, with all this, of course I had to give a distinction for this problem in view of very complex technical achievement in this theme and with such type or twins. Besides that, there is an amusing moment that united all three phases together: at the beginning – in diagram position – wK is stalemated on 8th rank and in the end – in final position of c) – on 1st!
EN <-> RU

Special Honorable mention - No 131, Solaiappan Manikumar (India) 2B5/p1ppK3/P1P4p/4NknR/3P4/2N1B3/2rP1b2/1Q3R2

1.Bb7? – 2.Bc8=, 1...d5 2.Ba8=, 1...dxc6 2.Bxc6=, 1...d6!
1.Rfh1? – 2.Rf1=, 1...Bxe3 2.dxe3=, 1...Be1 2.Rxe1=, 1...Bg1 2.Rxg1=, 1...Bh4 2.R1xh4=, 1...Bg3!
1.Qa1? – 2.Qb1=, 1...Ra2 2.Qxa2=, 1...Rxd2 2.Bxd2=, 1...Rc1 2.Qxc1=, 1...Rb2 2.Qxb2=, 1...Rxc3!
1.Rxh6! – 2.Rh5=
1...Se4 2.Qxc2=
1...Se6 2.Bxd7=
1...Sf3 2.Rxf2=
1...Sh3 2.Rxh3=
1...Sh7 2.Rxh7=
1...Sf7 2.Kxf7=
Compared to pdb/P1339309, the author have added 4th pin and corresponding 3rd try and 3rd stalemate with pin in solution. The fact of task deserves an HM but, in view of great resemblance to pointed partial anticipation, the distinction is only special.
EN <-> RU
Commendation - No 1
Gennadi Chumakov
TT-240, SuperProblem, 09-07-2020
Commendation - No 9
Aleksandr Tyunin
TT-240, SuperProblem, 09-07-2020
Commendation - No 45
Sébastien Luce
TT-240, SuperProblem, 09-07-2020

Commendation - No 1, Gennadi Chumakov (Russia) 6k1/3p2Pb/3p1Kp1/3P2P1/8/8/8/8

1...Ke7 2.Kxg7 Kxd7 (2...Ke8? 3.Kh8 Kf7 4.Bg8+ Kxg6 5.Bf7+ Kxf7 6.?? g6=) 3.Bg8 Kxd6 4.Be6 dxe6 5.Kf8 e7+ 6.Ke8 Ke6=
Solution is supplemented by try with another stalemate position for achievement of which White has no a tempomove.
EN <-> RU

Commendation - No 9, Aleksandr Tyunin (Russia) 8/3p4/2pN1p2/2p2p2/2P2p2/4pK2/4N2p/7k

1.Sd~? d5!
1.Sc8! d5 2.Sb6 d4 3.Sa4 d3 4.Sac3 dxe2 5.Sxe2=, 4...d2 5.Sd1=
The problem reminded me a little… famous study of Sarychev brothers! White piece (here – Khight, in Sarychev’s study – King) at first moves behind Black Pawn in order to “push” it and after that quickly catching up it. Even first move is coincidentally executed on the same square – c8!
EN <-> RU

Commendation - No 45, Sébastien Luce (France) 8/6p1/p2p2P1/p2P1p2/PpK1kPp1/1P2P1p1/3P2P1/8

1...d3+! 2.Kxe3 d4 3.Kxf4 Kd3 4.Kg5 Ke3 5.Kxg6! Kf4 6.Kh5 Kxf5 7.Kh4 Kg6=
Unusual beginning of the play: wP execute single (not double) move because otherwise after 1.d4? Kxe3 White (!) will be stalemated! Further it can be noted an annihilation of wPg6 for releasing a square.
EN <-> RU
Commendation - No 53
Dieter Werner
TT-240, SuperProblem, 09-07-2020
Commendation - No 58
Luis Miguel Martin
TT-240, SuperProblem, 09-07-2020
h=2.55 sol.(10+7)
Commendation - No 75
Marko Klasinc
TT-240, SuperProblem, 09-07-2020

Commendation - No 53, Dieter Werner (Switzerland) 8/8/8/8/3p1p1p/2pP1Ppk/2P3rp/5B1K

1...Be2 2.Rf2 Bd1 (Bf1+?) 3.Rd2 Be2 4.Rxd3 Bd1 5.Rd2 Be2 6.d3 Bxd3 7.Rg2 Bf1=
White and Black jointly helped each other to get desired tempo to pass a turn move to Black.
EN <-> RU

Commendation - No 58, Luis Miguel Martin (Spain) 8/8/7p/pPpkPP1p/K1n2P1P/pB4N1/P1N5/8

1...Sxa3 2.Kd4 Sc2+ 3.Kd4-d5 a3=
1...Sxh5 2.Ke4 Sg3+ 3.Ke4-d5 h5=
1...b6 2.Kc6 b7 3.Kc6-d5 b8S=
1...e6 2.Kd6 e7 3.Kd6-d5 e8S=
1...f6 2.Ke6 f7 3.Ke6-d5 f8S=
I will express a arguable opinion – I think HOTF with different quantities of solution looks harmonically only when such different quantities is justified by a concept. For example, on probation, HOT 3+4, where 3 is Valladao and 4 is AUW (albino, pickaninny and etc.).
For this reason the problem looks unfinished, that prevented me from getting Prize distinction for it. However even seemingly harmonic HOTF 2+2 with bK star will look boring here in view of full symmetry.
Apparently, due to big huge technical difficulties, here HOTF 4+4 with bK switchback bK rosette is unreachable…
That is why it remains to evaluate what is. And in any case, maximal quantity of phases (among all sent problems) – five – gives this problem a right for quite high distinction, after all Prizes at once.
Initially the problem was marked higher but the anticipation yacpdb/345036 with 8 (!) thematic solutions was discovered. But in that problem all solutions are quite heterogeneous, while here all 5 solutions are united by bK play with switchbacks – it is allowed the problem to remain in the award.
EN <-> RU

Commendation - No 75, Marko Klasinc (Slovenia) 8/8/8/8/pp6/kp1p4/rn1P4/bKB5

1...Bxb2+ 2.Rxb2+ (2.Bxb2? pat!) 2...Kc1 3.Rc2+ Kd1 4.Rc3 dxc3 5.d2 Kxd2 6.b2 Kc2 7.b3+ Kb1=
Rundlauf of White King.
EN <-> RU
Commendation - No 107
Terho Marlo
TT-240, SuperProblem, 09-07-2020
Commendation - No 118
Alberto Armeni
TT-240, SuperProblem, 09-07-2020
Commendation - No 141
János Csák & Gábor Tar
TT-240, SuperProblem, 09-07-2020

Commendation - No 107, Terho Marlo (Finland) 8/8/8/4p1p1/4P1P1/4K1pB/4b1P1/4k3

1...Kf1 2.Kd2 Kg1 3.Ke3 Bf1 4.Kf3 Bxg2+ 5.Kxg3 Bh1 6.Bg2 Bxg2=
Compared to diagram position, in the final both Kings and Black Bishop “shift” on two verticals righter – on the line “g” – and form an echo-stalemate compare to a diagram.
EN <-> RU

Commendation - No 118, Alberto Armeni (Italy) 5k2/1p1p1P1p/1P1P1K1P/2P5/5p2/5P2/8/8

1...bxc6 2.b7 c5 3.b8S c4 4.Sa6 c3 5.Sb4 c2 6.Sxc2=
1...dxc6 2.d7 c5 3.d8B c4 4.Bc7 c3 5.Bxf4 c2 6.Bc1=
Two underpomotions.
EN <-> RU

Commendation - No 141, János Csák & Gábor Tar (Hungary) 5brr/4p1pk/4P1p1/6P1/7P/1p6/pP6/K7

1.h5 2.h6 3.hxg7 4.gxh8S 5.Sxg6 6.Sxe7 7.Sxg8 8.e7 9.exf8Q 10.Se7=
Amusing concept: in diagram position a “clew” from 6 pieces is “attached” to bK, in solution White captures all 6 these pieces and create a new stalemate!
EN <-> RU

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