Final Award in Quick Composing TT-184 | Окончательные итоги блицконкурса TT-184

Rebirth square paradox | Парадокс поля возрождения

Theme | Тема

41 entries were received from 19 authors representing 11 countries | На конкурс поступило 49 композиций от 19 авторов из 11 стран

EN <-> RU

For this award 41 problems were submitted to me on plain, unnamed diagrams in an identical format. Ten of these entries failed to meet the theme specification: No 4A (Kc1-Kh7), No 4B (Kd5-Kh7), No 5 (Kh6 Ka6), No 6 (Ke1-Kd3), No 7 (Ka6-Kg7), No 9 (Ka6-Kc6), No 11 (Ke8-Kd4), No 12 (Kg7-Kb4), No 16 (Ka7-Kg5), No 20 (Kh6-Kf6), No 21 (_-a8). Required was that (using appropriate fairy rebirth conditions) White in the case of mate stipulations or Black in the case of selfmate stipulations should paradoxically profit from the fact that one of his units is not reborn but removed completely from the board, with consequent loss of both its mass and power. Similarly stalemate and selfstalemate were also allowed under the same conditions.

More than half of the entries bore the stipulation h#2 Circe but in almost all cases the authors of these problems had made things easy (or too easy!) for themselves by using simple schemes in the manner of the 1970s, tailored to the set theme in a purely formal way, and not presenting any new paradoxical effects in the spirit of the tourney requirement. The same may be said of the single #2 Circe [No 8 (Kh7-Kf1)]: it makes do with a Dombrovskis-effect which has often been seen before. In contrast the three prizewinners have achieved novel, imaginative settings of the theme by using skillful combinations of fairy elements.

Before I come to the details of the award here are a few more notes on entries which had to be eliminated after a single examination:
- No 1 (Kb4-Ka7). Not thematically original enough;
- No 10 (Kb6-Kc2), No 14 (Kd6-Ke2), No 32 (Kf8-Ke4) and No 34 (Ka8-Kd3). By the very fact of their symmetrical structures these essentially contain only a single thematic line;
- No 29 (Kc5-Ke4). With its ser-#2 Circe stipulation this is more a constructional sketch than a finished problem;
- No 31 (Kc1-Ka4). 25 units, of which 9 are unorthodox pieces, is too much material in proportion to the theme;
- No 36A (Kc6-Kh4), No 36B (Kd6-Kh4). Madrasi is not thematically necessary, serving merely to allow the WK access to g6;
- No 41 (Kd7-Ka8). A single dummy is fundamentally just a constructional makeshift; eight (!) of them are enough to exclude any tourney distinction.

I´ld like to thank Chris Feather for his translation of the award from German into English.

Award is the following | Отличия распределились следующим образом

1st Prize, 1st Place - No 38
Arnold Beine
TT-184, SuperProblem, 17-04-2017
1n3Q2/2rb4/3Np3/4k3/8/K1p5/B1B5/8
hs#2b) wSa2
Circe File, Annan Chess
(5+6)
2nd Prize, 2nd Place - No 33
Luis Miguel Martin
TT-184, SuperProblem, 17-04-2017
5r2/3K4/p1Pp4/R2kbp2/2pn4/8/7n/7B
h#22.1..
Provocation Chess
Mirror-Circe
(4+9)
3rd Prize, 3rd Place - No 17
Franz Pachl
TT-184, SuperProblem, 17-04-2017
4B3/P1nb3p/4N2k/2K5/4q3/7p/3pp2n/8
h#22.1..
Circe + Lortap
(4+9)

1st Prize, 1st Place - No 38, Arnold Beine (Germany) 1n3Q2/2rb4/3Np3/4k3/8/K1p5/B1B5/8

a) diagram: 1.Sc8! (1.S~? ... 3.Kxc5[+bBc8]! with Bc7) 1...Bc6 2.Qc5+ Bxc5[+wQc1]#
b) wSa2: 1.Qc8! (1.Q~? ... 3.Kxc4[+bSc8]! with Sc7) 1...Sc6 2.Sc4+ Sxc4[+wSc1]#
An airy Meredith position without white pawns. The ingenious combination of the two unorthodox conditions (exploited in both the black and the white play) gives rise to two striking solutions. A further point in the problem's favour is the characteristic Annan Chess twinning device, which changes the way in which the white king moves. However the crowning feature, which secures first place for this work, is the subtly paradoxical motivation for the failure of the two thematic try-lines. In each of the two phases White's first move must be an anticipatory occupation of c8, with the blocking pieces interchanging their functions. In both cases if that square were vacant a deeply-hidden defence would be available to the white king. Thus from a3, when equipped with the moving powers of the white bishop on a2 (in part a), he could capture the checking bishop on c5 because its rebirth on c8 would make the black rook on c7 move as a bishop, losing control of c5. An analogous effect would occur in part b, with the king on a3 having the power of the knight on a2 and thus able to capture the knight on c4, whose rebirth on c8 would once again cause the black rook to lose control, this time of c4.
EN <-> RU

2nd Prize, 2nd Place - No 33, Luis Miguel Martin (Spain) 5r2/3K4/p1Pp4/R2kbp2/2pn4/8/7n/7B

1.Rh8 (R~?) Rc5+ 2.dxc5 (dxc5[+wRh8]) Bf3# (3.Shxf3[+wBc8]!)
1.Rc8 (R~?) Be4+ 2.fxe4 (fxe4[+wBc8]) Rb5# (3.axb5[+wRa8]!)
In this case too, a combination of two unorthodox conditions makes the problem both original and prizeworthy. Factors determining its quality are the use of just two white pieces, sacrificing themselves alternately in Zilahi fashion and enabling black square-blocks, and the paradoxical refutations of the try-lines, which are wholly in the spirit of the tourney requirement. In accordance with the theme the rebirth squares (h8 & c8) of the two thematic white units must be blocked on the first move. After 1.R~? Rc5+ 2.dxc5[+bRh8]! Bf3+ Black has the reply 3.Shxf3! because this knight is enabled to capture by its observation by the white rook on h8; analogously after 1.R~? Be4+ 2.fxe4 [+bBc8]! Rb5+ there is 3.axb5! because the white bishop is observing a6.
EN <-> RU

3rd Prize, 3rd Place - No 17, Franz Pachl (Germany) 4B3/P1nb3p/4N2k/2K5/4q3/7p/3pp2n/8

1.Sf1 a8R 2.Qxa8[+wRh1] Rxh3#
1.Qb1 a8Q 2.Sxa8[+wQd1] Qxd2#
A further fine example of the exploitation of original and well-hidden motifs inherent in unorthodox conditions. Black's occupation of f1 or b1 (the original squares of the white bishop and knight) eliminates a Lortap-typical refutation of the intended mating moves. In the try-lines 1.Sg4? ... 2. ... Rxh3+3.BxBe8[+wBf1]! and 1.Qh1? ... 2. …. Qxd2+ 3.BxSe6[+wSb1]! the observing effect of the reborn white piece deprives a colleague (Rh3 or Qd2) of the power to check, a paradox which is perfectly in keeping with the intentions of this theme tourney. The economical use of material and the intensive exploitation of both unorthodox conditions also help to make this a worthy prizewinner.
EN <-> RU
1st Honorable mention - No 18
Franz Pachl
TT-184, SuperProblem, 17-04-2017
5rBK/3Rprp1/1p3p2/1P2B3/p7/knP5/8/3N4
h#2b) –Pa4
Circe
(7+9)
2nd Honorable mention - No 19
Igor Kochulov
TT-184, SuperProblem, 17-04-2017
1kq(B3)b3/(r3)(q3)2pP2/P3n3/2Q5/1KP5/1(B3)6/2n5/(R3)1(r3)5
h#22 sol.
Functionary Chess, Circe

a1, c1, a7: Pao (PA)
d8, b3: Vao (VA)
b7: Leo (LE)
(8+9)
3rd Honorable mention - No 37
Arnold Beine
TT-184, SuperProblem, 17-04-2017
pkb5/p7/2K1p3/2p1p3/2bPP3/1P6/RB6/8
h#2b) Ra2->a3
Circe File, Annanchess
(6+8)

1st Honorable mention - No 18, Franz Pachl (Germany) 5rBK/3Rprp1/1p3p2/1P2B3/p7/knP5/8/3N4

a) diagram: 1.Sa1 Rxe7 2.Rxe7 Bd6#
b) –Pa4: 1.Sc1 Bxf6 2.Rxf6 Ra7#
The best of the entries with the stipulation h#2 Circe but no further conditions or unorthodox pieces. The black knight on b3 and rook on f7 have to move away so as to open the bishop's doubly-closed diagonal from g8 to a2. The rook can achieve this only as a result of an active white sacrifice on e7 or f6, but the sacrificed pieces would be reborn with check from a1 or c1; therefore the black knight must start the play by occupying those squares while simultaneously opening the diagonal. Compared with the three prizewinners this problem is structurally less complex, but has enough good features to justify its place at the head of the rest of the field, being clearly constructed without blemishes, with good twinning and with diagonal/orthogonal correspondence in the two mates.
EN <-> RU

2nd Honorable mention - No 19, Igor Kochulov (Russia) 1kq(B3)b3/(r3)(q3)2pP2/P3n3/2Q5/1KP5/1(B3)6/2n5/(R3)1(r3)5

1.PAa5 LEa8 (1...LEc7?) 2.Qxc8 Kxc8[+wQd1]#
1.VAa5 LEc7
(1...LEa8?) 2.Qxa7 Kxa7[+wQd1]#
On the second move the white queen has the choice of capturing the black queen on c8 or the pao on a7 with check, so as to force the black king into a recapture which brings about selfmate because from the rebirth square d1 she observes the black knight on c2. In conformity with the thematic requirement White's first moves must take care to prevent the rebirth of the captured black units, for that would reduce the mating moves to harmless checks answered by any move of White Queen. An interesting way to set up the theme, but the congested position with its multiplicity of unorthodox pieces prevents its upgrading to the status of a prize.
EN <-> RU

3rd Honorable mention - No 37, Arnold Beine (Germany) pkb5/p7/2K1p3/2p1p3/2bPP3/1P6/RB6/8

a) diagram: 1.Bc4-c3 Ba3 2.Ba1! (2.B~? ... 3.Kxa7[+wBa1]! with 'B'a2) 2...Ba3xa7#
b) Ra2->a3: 1.cxd4[+wPd2] b3-a4 2.Ba2! (2.B~? ... 3.Kxa7[+wPa2]! with 'P'a3) 2...a4xa7#
Concentrated play on the a-file exploiting both unorthodox conditions but in a somewhat cramped position. The thematic content consists of the black bishop's occupation of a1 and a2 (the rebirth squares of the B and the P when they reach a7) so that the white rook (from a2 or a3) does not lose the power to guard that square. The award acknowledges the value of that idea, but both constructively and in its overall impression this work is outshone by the first prize problem, which is obviously by the same composer.
EN <-> RU
1st Commendation - No 26
Zoltán Laborczi
TT-184, SuperProblem, 17-04-2017
7q/6n1/B1K5/R7/1k6/8/8/7R
ser-h#6
Circe
(4+3)
2nd Commendation - No 23
Zoltán Laborczi
TT-184, SuperProblem, 17-04-2017
8/7B/8/8/5p2/3PBkN1/4n3/4K3
h#2b) Ke1->h1
Circe
(5+3)
3rd Commendation - No 30
Zoltán Laborczi & Gábor Tar
TT-184, SuperProblem, 17-04-2017
8/4(B3)1p1/Pb4(R3)1/k1p3p1/3P4/ppN1p3/4p3/6K1
ser-#32.1..  Circe
b) without Circe

e7: Vao (VA)
g6: Pao (PA)
(6+9)

1st Commendation - No 26, Zoltán Laborczi (Hungary) 7q/6n1/B1K5/R7/1k6/8/8/7R

1.Se6! 2.Qa1! 3.Kxa5 4.Qf1! 5.Kxa6 6.Qf7! Ra1#
A little jewel which owes its sparkle to the artistically elegant setting.
EN <-> RU

2nd Commendation - No 23, Zoltán Laborczi & Gábor Tar (Hungary) 8/7B/8/8/5p2/3PBkN1/4n3/4K3

a) diagram: 1.Sc1! Be4+ 2.Kxe3 Sf5#
b) Ke1->h1: 1.Sg1! Bg6 2.fxg3 Bh5#
In both phases Black occupies the rebirth square of a white piece (Be3 or Sg3) with his knight, so that his king may make a capture without incurring a check by rebirth. The airy, economical position and the model mates make this eight-piece problem stand out above a number of similar examples.
EN <-> RU

3rd Commendation - No 30, Zoltán Laborczi (Hungary) 8/4(B3)1p1/Pb4(R3)1/k1p3p1/3P4/ppN1p3/4p3/6K1

a) Circe:
1.PAe6! 2.PAxe2 3.PAa2# [3...bxa2[+bPAa8]? selfcheck]
1.PAf6? 2.PAf1 3.PAa1+ Kxa6[+wPa2]!
1.VAxa3[+bPa7]! 2.VAc1 3.VAd2# [3.exd2[+bVAd8]? selfcheck]
1.VAd6? 2.VAg3 3.VAe1+? cxd4[+wPd2]!
b) without Circe:
1.PAf6! 2.PAf1 3.PAa1#
1.VAd6! 2.VAg3 3.VAe1#
The theme is limited to the role of the two white pawns on a6 and d4, because the stipulation itself renounces the possibility of any black play. After a series of three moves the pao checks from both a1 or a2 or the vao from e1 or d2, but only 3.PAa2 and 3.VAd2 give mate since (as required by the theme) the latter moves prevent the replies 3…Kxa6[+wPa2] and 3...cxd4[+wPd2], Further, the attempts to avoid mate by the captures 3…bxa2[+bPAa8] and 3...exd2[VAd8] fail because of selfcheck from the reborn pieces. In an addition to the theme itself the problem is rounded off by the original twin using orthodox play ending in 3.PAa1 and 3.VAe1, which are the only possibilities now that all rebirths have been excluded along with the fairy condition.
EN <-> RU
4th Commendation - No 39
Ivo Tominić
TT-184, SuperProblem, 17-04-2017
1N6/PrRpN3/1P3p2/3n4/8/P6K/2PP4/1BB4k
h#22.1..
Circe
(11+5)
5th Commendation - No 22
Ralf Krätschmer
TT-184, SuperProblem, 17-04-2017
7K/4p3/4Ppp1/2PkP3/2pNp3/8/8/B2B4
h#2.5b) Ba1->g3
Circe
(7+6)

4th Commendation - No 39, Ivo Tominić (Croatia) 1N6/PrRpN3/1P3p2/3n4/8/P6K/2PP4/1BB4k

1.Sxc7[+wRa1] Ba2! (A) 2.Rxa7 Bb2# (B) (1...Bb2? 2.Rxa7[+wPa2] Ba2??)
1.Rxc7[+wRa1] Bb2! (B) 2.Sxb6 Ba2# (A) (1...Ba2? 2.Sxb6[+wPb2] Bb2??)
A pretty AB/BA choice by the two front pieces of a half-battery set up by the first moves. The neat mechanism combining the black rook and knight might have deserved an Hon. Mention but for the two static white knights, which remain idle even in the mates.
EN <-> RU

5th Commendation - No 22, Ralf Krätschmer (Germany) 7K/4p3/4Ppp1/2PkP3/2pNp3/8/8/B2B4

a) diagram: 1...Be2 2.Kxe5 Bxc4[+bPc7] 3.f5 Se2#
b) Ba1->g3: 1...Se2 2.Kxe6 Bf4 3.Kf5 Sd4#
White's first moves occupy the thematic square e2 with the bishop or knight, so that there can be no rebirth when the black king captures one of the e-pawns. The theme is achieved; however the harmful effects in the corresponding try-lines (if there is a white pawn on e2) are unfortunately not well matched. In part a) the pawn prevents the mate Se2 because it occupies that square, while in b) it closes the diagonal from d1 to h5, giving the black king a flight on g4. The twinning is not ideal.
EN <-> RU


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f# (fairies | сказки)

Participants | Участники

Bašić B. – No 41
Beine A. – No 37, 38
Caillaud M. – No 3
Ganapathi G. – No 21
Gockel H. – No 8
Golha J. – No 2, 4A*, 4B*
Kochulov I. – No 19
Krätschmer R. – No 22
Kuhn R. – No 15*, 20, 32*, 40
Laborczi Z. – No 23, 24, 25, 26, 28*, 29*, 30*
Lind I. – No 34, 35, 36A, 36B
Martin L. M. – No 33
Müller D. – No 9, 10, 11, 12, 13, 14*, 15*, 16, 32*
Pachl F. – No 14*, 17, 18
Packa L. – No 1
Skoba I. – No 4A*, 4B*
Stepochkin A. – No 5, 6, 7
Tar G. – No 27, 28*, 29*, 30*, 31
Tominić I. – No 39

The Winner | Победитель

Arnold Beine
Congratulations! | Поздравляем!

Judge | Арбитр

Klaus Wenda

Director and editor
Директор и редактор

Aleksey Oganesjan
e-mail: alexeioganesyan@gmail.com

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