онкурс посв¤щаетс¤ пам¤ти —ерге¤ Ўеде¤

 »тоги блицконкурса ““-66 Ђ#2; #3-5 с элементами ретроанализаї.

Award, Section #2

 A total of 25 entries from 16 authors were received.

9 problems were cooked. That is a high percentage! As regards soundness modern composers have grown accustomed to relying on the computer; so when for some reasons this becomes impossible, they start experiencing difficulties.

Among the sound entries, the following ones appear in the award.

Ќа  конкурс  поступило  25  задач  16  авторов.

9  задач  дефектны.  Ѕольшой  процент!  —овременные  композиторы  в  вопросах  корректности  привыкли  довер¤ть  компьютеру, и когда по  некоторым  причинам  это  становитс¤  невозможно,  начинают  возникать  сложности.

—реди  корректных  работ  отмечены  следующие:

 1^st Prize Ц 1^st place

The problem features Vladimirov theme, Rukhlis changes and Zagoruiko.

The key is flight-giving.

¬  задаче  есть  тема  ¬ладимирова,  перемены  по  –ухлису и «агоруйко. 

¬ступительным  тематическим  ходом  отдаетс¤  поле  черному  королю. 

2^nd Prize Ц 2^nd place

Full form of Vladimirov theme featuring the use of the en passant capture paradox

ѕолна¤  форма  темы  ¬ладимирова c использованием парадокса вз¤ти¤ на проходе.

3^rd Prize Ц 3^rd place

b) after WhiteТs first move

a.)  1. Qf2?  h2!

What was the last move?

0Е...c6:b5? Ц impossible, since the 16^th white piece, i.e. the one that is missing, is the Bc1, which cannot be captured on b5.

Of the two last moves of seemingly equal value Ц 0. Еd7-d5  and  0. Еf7-f5 Ц the former is impossible, since in that case the Ba6 is promoted, while the promotion squares of the missing black pawns are dark.

Therefore, the only possible last move is
0Е.f7-f5!   Then 1.g:f6! e.p. g:f6+ 2.S:f6#

          b.)   Now 0Е.d7-d5! is possible, because the Ba6 could have been promoted on f1! Solution: 1.c:d6! e.p.  g:f6+  2.S:f6#

 This problem is the most interesting one in the tourney as regards retro reasoning. Two e.p. captures; remarkably, the second twin presents a Уblock problemФ in which the e.p. capture is the only waiting move. However, the theme required focusing on the orthodox component of the entries; hence 3^rd place.

ƒанна¤  задача  сама¤  интересна¤  в  конкурсе  по  ретрообоснованию. ƒва  вз¤ти¤  на  проходе,  причем  во  втором  близнеце Ц  задача-блок,  вз¤тие  на  проходе  Ц  единственный  выжидательный  ход.  Ќо  в  за¤вленной  теме  упор  делаетс¤  на  ортодоксальную  составл¤ющую  задачи,  отсюда  и  третье  место.

Special Prize for a rendition of S. ShedeiТs idea Ц 4th place

   b) after the first moves by White and Black

а) 1.0-0-0?+  Kc3!  (2.Rd3+? Kc4!)

    1. h8Q+!  e5   2.de6 e.p.#

    1.h8B+!?  e5  2.de6 e.p.#??

The second solution of the first twin (in this problem, promotion to bishop) is ruled out because it would lead to no solution in the second twin Ц this complex Уextended problemФ idea was proposed by S. Shedei (see Shakhmaty, 1974, Masters of Chess Composition, p. 77, No. 196). The problem can also be found in TurevskyТs database; the link is  http://www.yacpdb.org/?id=295923 

b) 1.de6# e.p.?? Ц the e.p. capture is illegal!

    1.Q:e5+?  K:e5  2.Bb2#,  but 1Е.de5!

    1.Rc1?    a1R  2.Bb2#     1Е.a1B  Rc4#,    but 1Е.a1Q!

    1.0-0-0+!  Kc3  2.Qh3#

Valladao task involving underpromotions for both sides in a twomover. The problem demonstrates the idea of mathematical paradox Ц when 2 Ц 1 = 2 Ц which is seldom used in forward-play problems

“аск ¬алладао с использованием слабых превращений обеих сторон в двухходовке.

¬  задаче  продемонстрирована  редко  используема¤  в задачах с пр¤мой игрой ретроиде¤  математического  парадокса,  когда  2-1=2.

1^st Honorable Mention Ц 5^th place

With the help of a retro element, the problem presents a combination of black rookТs УcrossФ with Hannelius and change of white movesТ functions.

ѕри помощи ретроэлемента в задаче реализовано сочетание крестика черной ладьи с темой ’аннелиуса и переменой функций ходов белых.

2^nd Honorable Mention Ц 6^th place

1.Rb7? [2.Ra7#], __ 1ЕRxa5#! Ц the only black defense results in mate to the white king.

1.Rxd5? zz [2.Qa7#], __ 1Еcxd6!

1.Qxc5? [2.Qa7#], __ 1Еg1=Q/B!

1.dxc7? [2.Sb6#] 1ЕRxb5/Rxc6 2.Qa7#,   but Black is to move.

 0Еcxd6 1.Rxd5 zz 1Е~ 2.Qa7#

0Еd4 1.Rb7! [2.Ra7#] 1ЕRxa5+ 2.Rxa5# Ц now itТs not mate, but just a check to the white king.

1.dxc7? , __ 1ЕRxb5/Rxc6!

Berlin theme Ц rare for a two-mover.

 Ѕерлинска¤  тема Ц  редка¤  дл¤  двухходовки.

Special Honorable Mention for a scacchographical problem Ц 7^th place

УThe Commemorative LampФ

Change of play in a scacchographical retro problem with a retroanalytically motivated refutation of the try.

ѕеремена игры в скахографической ретрозадаче с ретроопровержением ложного следа.

1^st Commendation Ц 8^th place

1.Qa1?  -  2.fg#

1Е.gf   2.Q:f6#,    1Е.g6(g5)!

1.Qf8? - 2.Q:g7#(fg#)  1Е.gf!

1.B:f7+  K:h7!,

but Black is on the move.

0Е.g5(g6)  1.Qf8  Ц  2.Q:g7#

0Е.gf         1.Qa1  Ц  2.Q:f6#

0Е.b5       1.B:f7+   Kh7  2.Qh1#

b)

1.K:f7?  -  2.fg#

1Е.gf    2.Q:f6#,   1,Еg6(g5)!

1.fg+  Kg7   2.Qf6#,  but again Black is on the move; therefore

0Е.g5   1.Q:g5  -  2.Qg7#

0Е.g6   1.Qh6  -  2.Qg7#

0Еb5   1.fg7+  K:g7  2.Qf6#

Change of play, reciprocal change of mates and first moves; yet itТs sort of boringЕ

ѕеремена  игры,  чередование  матов и вступительных  ходов,  но  как-то  скучноЕ

2^nd Commendation Ц 9^th place

A Meredith with retro-pickaninny.

I am afraid this has already been presented, even with a retro element; but I cannot point to a particular anticipation.

ћередит с ретропикенинни.

Ѕоюсь, что это уже было,  даже с ретроэлементом, но не могу представить ничего конкретного.

3^rd Commendation Ц 10^th place

Urania and Salazar themes.

I would like to point to the fact that the try is based on a retroplay that is in fact non-existent. The problemТs retro component is sort of virtual.

“емы ”раниа и —алазара.

’очу отметить идею использовани¤ несуществующей фактически ретроигры в качестве ложного следа. –етросоставл¤юща¤ задачи как бы  иллюзорна.

јрбитр ћарк Ѕасистый

******************************************************

Award in TT-66, section #3-5

When the tourney became a memorial one (alasЕ) and its deadline was extended, more entries started coming in, as composers paid tribute to the memory of outstanding master Sergei Shedei (thanks a lot to all of the participants!); as a result, the judge of the #3-5 section received a total of 25 problems from 12 authors representing 6 countries; 7 entries were unsound and 1 was non-thematic.

The award is as follows.

ѕосле того, как конкурс (увыЕ) стал мемориальным и были продлены его сроки, композиторы активизировались, отдава¤ дань пам¤ти выдающегос¤ мастера —ерге¤ Ўеде¤ (за что им огромное спасибо!), и в итоге по разделу #3-5 было получено 25 задач 12 авторов из 6 стран. 7 из них оказались некорректными, 1 Ц нетематической.

The problem features an idea proposed by Valery Liskovets back in 1977: combining partial retroanalysis (pRA) with the Уtouch Ц moveФ rule (usually designated as BG, from the German phrase УBerührt Ц geführtФ). Although this idea allows presenting change of play(mates)/
refutations within the framework of the Уpartial solutions,Ф composers have not used these opportunities in the past.

Retroanalysis shows that BlackТs last move was either c7-c5 or e7-e5. White starts with 1.fxe6+ e.p.!, implying that Black played 0Еe7-e5. 1ЕKxd8 2.e7+ Kd7 3.b6# (2.Rf8? ~ 3.Qa5# Ц 2Еc4! (a))

If at this point Black declares that in reality he played 0Еc7-c5?, it means that White performed an illegal move; instead of it, now he must make another move by the same piece of his own color or capture the same black piece. However, the white f5-pawn has no other moves to make and the black e5-pawn is Уuntouchable.Ф In such a case, according to FIDE rules, WhiteТs УimpossibleФ move (1.fxe6+ e.p.) is to be replaced with any other legal move. White chooses 1.bxc6+! e.p. (0... c7-c5) Kxd8 2.Rxd6+ Sxd6 3.c7# (model mate) (2.Rf8? ~ 3.c7# Ц 2ЕBb6! (b)

If White starts 1.bxc6+ e.p.? (0... c7-c5) Ц 1ЕKxd8 2.Rxd6+ Sxd6 3.c7#, then Black declares that in reality he played 0Еe7-e5 Ц and then, according to FIDE rules, White is forced to make a move with his b5-pawn: 1.b6+ (we have already seen this move Ц as mate after 1.fxe6) Kxd8 2.Ke6 [3.Qd7#] Sc7+ 3.bxc7# Ц 2...Sxf6! (2.Rf8? ~ 3. Q/R:e8# Ц 2ЕBxg6! (c))

 b) in the position before
BlackТs last move, White to play

1.Bxb7+! Kxb7 2.c6+ Ka8 3.b7+ Ka7 4.Bc5#

b) Black retracts 0-0.

1.Se5! ~ 2.Bxf7+ Kd8 3.Bc7+ Ke7 4.Sg6# (model mate); 1...Kd8 2.Bxf7 Kc8 3.Bc7! ~ 4.Be6# (same 2^nd and 3^rd moves by White, but without checks)

Well, how about castling? In the position of twin b), it is Black that made the previous move; he had to move his king or rook, thus losing the right to castle. The judge has already seen a twin formed according to the same principle (Black retracts, yet White moves); but that problem did not present the seemingly paradoxical idea of УlosingФ the right to castle after УretractingФ casting.

ј как же рокировка? ¬ позиции близнеца последний ход делали черные Ц королем или ладьей, и право на рокировку потер¤но. Ѕлизнец, который образуетс¤ подобным образом, судье уже доводилось видеть, но без внешне парадоксальной идеи Ђутратыї права на рокировку после ее Ђвозвращени¤ї.

a) BlackТs previous move c6xb5 or d6xc5 results in a disbalance for White: 11 (pieces on the board) + 4 (f7xe6xd5xc4, axb) +2 (bxc, c6xb5 or c7xd6xc5) = 17. If 0Еb7-b5, then the Ba8 is a promotee, resulting in a disbalance again, this time for Black: 9 + 6 (bxa, d2xc3, fxexd, gxfxe) + 2 (Bc8, Bf8) = 17. The moves 0Еb6-b5, b6xc5, Sc6-(x)c8 could not have been played because of illegal check to the white king; c6-c5, Kb7-(x)a7 are unretractable because of illegal check to the black king. Only 0Еc7-c5! remains, preceded by Rb6xg6+ (uncapture is a must, in view of the need to avoid retrostalemate for Black, since the retraction that seems to have become available Ц 1Ес5-с4 Ц is illegal: the c4-pawn came from the f7-square).

1.dxc6 e.p.+! Kxa8 2.c7! ~ 3.c8=Q ~ 4.Qb7# (2ЕSc6+ 3.Rxc6 ~ 4.c8=Q#)

b) Now Black could have also played 0Еb7-b5, WhiteТs promotion to bishop on a8 is possible, since the f8-bishop could have been captured by one of the white pawns. Therefore, 1.dxc6 e.p.+ is illegal. Now the solution is 1.Bxc5+! Kxa8 2.Rg8! ~ 3.a7! Kb7 4.axb8=Q#. The twins present a rather coherent and fairly original idea: the Уun-incarcerationФ of one of the black bishops makes it possible to УincarcerateФ the other one.

Content: retro-Avanta (УparallelФ double moves by two black pawns in retroplay) in synthesis with promotions of two white pawns that are adjacent to the black pawns; reciprocal change of functions of these white pawns (mate upon promotion Ц guarding a black kingТs flight square).

—одержание: ретро-јванта (Ђпараллельныеї ходы черных пешек на два пол¤ в ретроигре) в синтезе с превращени¤ми двух соседних с ними белых пешек; чередование функций этих белых пешек (мат после превращени¤ Ц подхват пол¤ возле черного корол¤).

4^th Prize Ц 4^th place No. 19, Igor Agapov

1.Sf5?! Kxb2 2.Sd4! Ka2 3.Sab5! Kb2 4.Rb3+ Ka2 5.Sc3#

BlackТs last move? Ka1(b1)-a2, preceded by 0-0+? But the castling is illegal, as the white king had to move to let his black counterpart Уwalk by.Ф Ka1(b1)xSa2, preceded by Sc1-a2+? But both white knights and all 8 white pawns are on the board. Ka1xBa2, preceded by Bb1-a2+?  But then it is retrostalemate for Black. And if Bb1xa2, then it is a disbalance for Black, since all 12 missing black pieces were captured by white pawns. So Black is on the move.

0ЕKxb2. Now whiteТs previous plan is refuted by stalemating: 1.Sf5? (1...Ka2 2.Sd4 Kb2 3.Rb3+! Ka2 4.Rb5! Kxa3 5.Ra1#) Ц

1...Kxc2! 2.Sd4+ Kd2!! 3. Sab5 Ц stalemate!

The solution is changed: 1.Rb3+! Kxc2 2.Rfb1! Kd2 3.R1b2+! Ц and now:
3...Kc1/e1 4.Ra2! Ц zz 4...Kd1 5.Rb1#;

3...Kd1 4.Rc3!  zz 4...Ke1 5.Rc1#

The two white knights do not take part in the solution, but, considering their role in the try and in view of the theme of the tourney, they clearly Уdo not spoil the race.Ф It is quite interesting that, according to the author, this composition is a correction of a problem he composed at the age of 13.

’оть два белых кон¤ в решении не участвуют, но с учетом их роли в ложном следе и темы конкурса они ¤вно Уборозды не порт¤тФ. »нтересно, что, по сообщению автора, эта композици¤ ¤вл¤етс¤ исправлением его задачи, составленной в возрасте 13 лет.

Special Prize Ц 5^th place  No. 3, Joaquim Crusats

One capture for White is obvious: gxh; the black h4-pawn is УbelowФ two white pawns; the УsecondФ white pawn could not Уpass byФ this black pawn, as this would require two more captures, while White has only one in store. This means that Black made two captures on the g- and h-files. WhiteТs balance is closed: 11 + 5 (a7xb6, bxc, cxd, hxg, gxh) = 16. The white d-pawn was captured on its original file by the black d4-pawn, while the white e-pawn was promoted on e8 after e6xd7. The d7-square is a light one and so the black dark-squared bishop could only have been captured on the h-file. The white a-pawn was promoted on a8; therefore, BlackТs previous move could not have been a7xb6. If so, how else can Black avoid retrostalemate for White? 0ЕB~-g4 1.g4xh5?? Ц the black dark-squared bishop could not have been captured on the light h5-square. 0Еg6-g5? 1.g5xBh6 Rg7-h7+ and retrostalemate for White. The white king has to be shielded: 0Еg7-g5!! 1.g5xBh6 B~-g4 2.g4-g5, etc. Hence White can capture en passant: 1.hxg6 e.p.! (no check!) ~ 2.Sxf7+ 2.Rxf7+ 3.Kxf7#; 1...Sb/dxc7/ Qxd7 2.Sxf7+ Rxf7+ 3.Kxf7+ S/Qxe8(+) 4.Rxe8 #; 1...Sf6 2.Sxf7+ Rxf7+ 3.Kxf7+ Sxe8/Sg8 4.Rxe8#/Rxg8#.

The judge has not seen the use of shielding to prove the legality of an e.p. capture. This problem is the one with the most profound retroanalysis in the tourney.

—удь¤ раньше не встречал использовани¤ экранировани¤ дл¤ обосновани¤ возможности вз¤ти¤ на проходе. —ама¤ глубока¤ в плане ретроанализа композици¤ конкурса.

1^st Honorable Mention Ц 6^th place No. 7, Dieter Müller

Black has no last move.

0ЕKxb8 (a) 1.Qxf7 Ka8 (switchback) 2.Qxc7 Bb8 (b) 3.Qb7#

0ЕBxb8 (b) 1.a7! Bxa7 (switchback) 2.Qa3 Kb8 (a) 3.Qf8#

An unusual reciprocal change Ц of BlackТs УzeroФ and second moves. A Meredith.

Ќеобычное чередование Ц УнулевогоФ и второго ходов черных. ћередит.

2^nd Honorable Mention Ц 7^th place  No. 16, Bosko Miloseski

1.exd5? f4!; 1.exf5? d4!; 1.Bf1? ~ 2.Qe2 ~ 3.Qb5/Qxa6#; 1...d4!;

1.Bxa6? ~ 2.Qe2 ~ 3.Qb5# ; 1...d4!; 1.Qe2? ~ 2.Bxa6 ~ 3.Qb5#; 1...d4!;

1.Rxf5? zz 1...d4 2.Kxd4 Kxb5 3.cxb6#; 1...Kxb5 2.Qe2+ Ka5 3.Qxa6#; 2...Kxc5 3.Se6#; 1...dxe4!

1.cxd6 e.p.! (0Еd7-d5) ~ 2.Rxf5 Kxb5 3.Kxd4#; 1...fxe4 2.Sxe4 Kxb5 3.Kd4# Ц in just five single moves, six (!) pieces УvanishedФ between the white queen and the black king; 1...f4 2.Sf3 Kxb5 3.Kd4#; 1...Kxb5 2.Qe2+ Ka5/Kc5 3.Qxa6#/Se6#

WhiteТs Уmasked battery consists of six pieces.

У«амаскированна¤ батаре¤Ф белых состоит из шести фигур!

3^rd Honorable mention Ц 8^th place No. 4, Grigory Popov

The first variant features white piece switchback; the second, УdoubleФ zugzwang.

ѕервый вариант Ц с возвратом белой фигуры, второй Ц с УудвоеннымФ цугцвангом.

1^st Commendation Ц 9^th place No. 12, Anatoly Vasilenko

The problem took part in a tourney requiring correction of cooked problems that was conducted on January 5, 2011, on the website ru-chess-art. It is based on a 1928 composition by G. Kasparyan, but the Уcorrection tourneyФ judge concluded that it was in fact not a correction, as it does not share a single common variant with the said problem, and so it can be regarded as a fairly original work.

«адача участвовала в конкурсе исправлени¤ дефектных задач на сайте ru-chess-art 5/01/2011. —оставлена по мотивам задачи √.  аспар¤на 1928 года, но, по заключению судьи конкурса, не ¤вл¤етс¤ исправлением, и, поскольку эта композици¤ не имеет с указанной задачей ни одного общего варианта, ее можно считать вполне оригинальным произведением.

2^nd Commendation Ц 10^th place No. 6, Anatoly Vasilenko

A miniature. The first couple of twins Ц thematic ones Ц present change of promotion; the second couple, which is non-thematic, presents unstalemating of the black king. All mates are different.

ћиниатюра. ѕерва¤ пара близнецов Ц тематическа¤, с переменой превращений; втора¤ пара Ц нетематическа¤, с распатованием черного корол¤. ¬се маты разные.

3^rd Commendation Ц 11^th place No. 21, Ralph Krätschmer

1.Sg6? Rxg6!
1.Bd6 ~ 2.Rxe7+ Kd8 3.Kf7#; 1Еexd6 2.Sg6! Rxg6/Bxg6 3.Kxh7#/Kxg7#. A nice Nowotny on the 2^nd move.

1Е0-0-0? Ц the castling is illegal, as the black king had to move to let the white Rh8 and Kg8 go to the upper right corner.

4^th Commendation Ц 12^th place No. 3, Grigory Popov

1. Bb6! Castling would save Black, but it is illegal. Either the king or the Rh8 had to make the last move; 0Еb7xa6, 0Еe7-e5, 0Еe7xf6 could not have been played on account of disbalance for Black. 1Еaxb6 2.Qd1 ~ 3.Qd5 ~ 4.Qa8#; 1Еg4 2.Sxe5+ Kf8 3.Sxd7+ Kg8 4.Bf7#

5^th Commendation Ц 13^th place No. 9, Mykola Cherniavsky

The only retro nuance consists in that the key move is made by a promoted white bishop.

≈динственный ретронюанс Ц первый ход делает белый превращенный слон.

Judge: Andriy Frolkin

Kiev, December 15-19, 2012

ѕобедители конкурсов - Dragan Stojnić - —¬ј ј „ј—“! и »горь јгапов - ѕоздравл¤ем!

19 декабр¤ 2012