No.1337 (AS)

No.1337 
Anatoly Stepochkin
(Russia)


(v) Anatoly Stepochkin & Jacques Rotenberg (Israel)
(v) Anatoly Stepochkin & René J. Millour
 (France)

Original Fairy problems
JF-2018/II:
July – December’2018


Definition: (click to show/hide)


No.1337 Anatoly Stepochkin
Russia

original – 21.10.2018

Solution: (click to show/hide)

white Kd8 Sc6e6 Pb2c2e4g2 black Kg4 Qg3 Rh3 Sh4h5 Pa2b4c7

#7                                               (7+8)
Maximummer


No.1337.1 Anatoly Stepochkin &
Jacques Rotenberg

Russia / Israel

version to No.1337 – 28.10.2018

Solution: (click to show/hide)

White Kc8 Bd7 Sf7g6 Pf5b2g2 Black Kg4 Qg3 Rh5h3 Ba8h2 Sf6b5 Pa7g7c6e5g5b4d4a2

#7                                               (7+16)
Maximummer


No.1337.2 Anatoly Stepochkin &
René J. Millour

Russia / France

version to No.1337 – 04.11.2018

Solution: (click to show/hide)

white Ka8 Rd4 Bh6 Pc2g2 black Kg4 Qg3 Rh3h5 Sa5c6 Pb2f4f5h7

#7                                              (5+10)
Maximummer


24 Responses to No.1337 (AS)

  1. Jacques Rotenberg says:

    Nice and neat. Would the author manage another white switchback instead of c2-c4-c5, it could be even better.

    • Georgy EvseevGeorgy Evseev says:

      It seems too easy to mate black king in so many moves, and double switchback seems impossible.
      At the same time, it is possible to implement multiple openings and closings of line b3-g8 with single white piece in a lighter construction. I am not sure that this is an improvement, though.

      For example: Kc7 Sc6g8 Pb2e4g2 – Kg4 Qg3 Rh3 Sh5 Pa2h4e5

      1.b3 Q*b3 2.Sge7 Qg8 3.Sd5 Ra3 4.Sf4 Qb3 5.Se6 Qh3 6.g3 R*g3 7.S*e5 #

  2. Joost de HeerJoost says:

    On the diagonal it can be done more economical, but the white switchback is missing.
    W: Ke8 Sf5 Pb2 Pg5
    B: Kg8 Qg7 Rh7 Bb7 Bh8 Sh5 Pa2 Pb4 Pc6
    #7, BlackMaximum

  3. Jacques Rotenberg says:

    Joost, you may – easily – have switchbacks with your idea :
    White : Ke8 Sd4 Pg6b2
    Black : Kg8 Qg7 Rh7 Bh8b7 Sh5 Pf3a2c2

    !.Sc6? Qxb2 2.g7 Qxg7 3.Se7+ Qxe7!
    1.S×c2!Q×b2 2.Sd4 Qh2 3.Sc6 Ba1 4.Se5 Qb2 5.Sc6 Qh8 6.g7 B×g7 7.Se7‡

  4. Anatoly Stepochkin says:

    George’s version is more economical, but the attempt was lost.

    Анатолий Стёпочкин.

  5. Jacques Rotenberg says:

    This is heavier, but it works :

    White : Kc8 Bd7 Sf7g6 Pf5b2g2
    Black : Kg4 Qg3 Rh5h3 Ba8h2 Sf6b5 Pa7g7c6e5g5b4d4a2

    I must thank Georgy who said : “…seems impossible..”
    It challenged me.
    It was not easy, indeed.

  6. Jacques Rotenberg says:

    1.b3! Q×b3 2.Sd8 Qg8 3.Be6 Ra3 4.Bd7 Qb3 5.Sf7 Qh3 6.g3 R×g3 7.Sf×e5‡

  7. Anatoly Stepochkin says:

    Position was hard, but the game is now flawless! I offered Yulia to publish it in collaboration Anatoly Stepochkin + Jacques Rotenberg. Jacques you don’t mind?

  8. Jacques Rotenberg says:

    Thank you Anatoly

  9. René J. Millour says:

    Adding 8 pieces to the 15 initial ones just for a second white switchback is highly questionable. From my point of view, this position is not only “heavier”, it appears particularly heavy.
    Suggestion in no more than 15 pieces, featuring bQbR after 2 white (and 1 black) switchbacks.
    White Ka8 Rd4 Bh6 Pc2 Pg2
    Black Kg4 Qg3 Rh3 Rh5 Sa5 Sc6 Pb2 Pf4 Pf5 Ph7
    1.Rxf4+? Qxf4!
    1.c3 Qxc3 2.Rd8 Qh8 3.Bg7 Ra3 4.Bh6 switchback Qc3 switchback 5.Rd4 switchback Qh3 6.g2 Rxg2 7.Rxf4#.

  10. Anatoly Stepochkin says:

    Rene, you’re right. We believe that the content of the above form, so went to such a difficult position. You found a lighter version using a different white material. This is exactly the case when one head is good and three is better. I think that You are not against the publication of the task in the triple Commonwealth Stepochkin+Rotenberg+Millour.

  11. René J. Millour says:

    Anatoly, thanks for adding my name but your message forces me to be clear.
    Apart from your nice initial matrix, the proposal by JR and my own proposal have nothing in common. From my previous message you can deduce I totally disapprove the heavy position in 23 pieces. Therefore, I will never accept a common version featuring 3 authors.
    My proposal: presenting in Julia’s Fairies the 3 successive versions.
    1) AS 15 pieces 2) AS+JR 23 pieces 3) AS+RJM 15 pieces. The judge will judge.
    This presentation according to the 3 steps that occurred is simply NATURAL

  12. Jacques Rotenberg says:

    Your version, Rene Millour, is very light and nice,
    Bravo !
    It seems that even you could also have a model mate :
    White : Ka8 Rd4 Bh6 Pc2g2
    Black : Kg4 Qg3 Rh3 Bf4 Sa5b5 Pc7h5b2d2

  13. Anatoly Stepochkin says:

    Rene, how Do you like the version of Rotenberga with the right Mat?

  14. René J. Millour says:

    I did not learn anything because I know “Bf4-Pc7-Pd2” for a while. I do not like c7 and d2, I do not like either f5 and h7 in what I sent. JR is now coming quickly with details. Model mate is simply an ornament, not at all the essence of my light position in 15 pieces. So, model mate or not could be your decision as first author.
    What is sure, I have reasons for that, I will never sign anything with JR. You have my proposal in my previous message.

  15. Jacques Rotenberg says:

    “…What is sure, I have reasons for that, I will never sign anything with JR…”

    This is a personal attack, so, I think, it requires explanations.

    By the way, in the meantime, I can only agree with Rene-Jean, that better is not to put my name besides his.

    And also, it fits well, because there is no necessity.

  16. Seetharamanseetharaman says:

    Rene Millour’s version is a masterly improvement of the original Well done!

  17. Jacques Rotenberg says:

    “…Adding 8 pieces to the 15 initial ones just for a second white switchback is highly questionable….” has been written. I think it is a little exaggerated.

    – Obviously, the theme shown requires switchbacks, so till the switchbacks are not shown, the idea is not complete, and it is pointless to start counting the pieces.

    moreover, to compare No.1337.2 with No.1337.1 may be interesting :
    1337.2 is (much) lighter, of course, this is a great advantage and it has to be seen as a clear improvement.

    It should be noted, however, that some of the positive aspects of 1337.1 are lost :

    -In 1337.1 the mate involves 3 officers, it is model and more complex
    -The switchback of the bishop is along its own line so that the return is made only for opening the line of the queen -thematic content.
    -The heavier white forces means that the absence of cooks is more a miracle than in 1337.2
    -The black lines are longer

  18. René J. Millour says:

    Many thanks, Anatoly, for this joint composition featuring a substantial content, now in a decent position.
    Also many thanks, Julia!

  19. Jacques Rotenberg says:

    Another “detail” :
    in 1337.1 the knight from d8 has a choice : it can play to f7 or to c6 to get to e5, and so the closing of the black line is active.
    In 1337.2 it is mechanical, the return to d4 by the rook d8 is the only way to f4.
    In a problem showing switchbacks, the quality of the switchbacks is an important feature.

  20. Jacques Rotenberg says:

    In fact, the more I compare 1337.1 and 1337.2, the more I like 1337.1.
    To appreciate a problem only by the number of pieces is highly questionable.
    The two versions are quite different, they both have their own interest.

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