No.612,612.1 (KS&LP)

ksoljaNo.612 
Kenneth Solja 
(Finland)

No.612.1
Kenneth Solja (Finland) &
Ladislav Packa (Slovakia)

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Original Problems, Julia’s Fairies – 2014 (III): September – December

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No.612 by Kenneth Solja – Three black promotions! (JV)

No.612.1 by Kenneth Solja & Ladislav Packa – A new version of No.612, created during a long discussion! One more half-move, one more solution and sacrifices of the white pieces! (JV)


Definitions:

Anti-Andernach: A piece (excluding King) changes its color after any non-capturing move. After capture, the piece retains its color. Rooks on a1, h1, a8 and h8 can be used for castling, provided the usual other rules for that move are satisfied. After castling, Rooks do not change color, If White makes a non-capturing move with neutral or halfneutral piece, that piece becomes black and vice versa.


No.612 Kenneth Solja
Finland

original – 04.10.2014

Solutions: (click to show/hide)

White Kc2 Rc7 Pe5 Pd4 Pg4 Pf3 Ph2 Black Ke6 Rh6 Sd1 Pd5 Ph5 Pb5 Pc3 Pe3 Pb2 Pf2 Pf4

h#2             3 solutions         (7+11)
Anti-Andernach

No.612.1 Kenneth Solja &
Ladislav Packa

Finland / Slovakia

Version of No.612 – 08.10.2014

Solutions: (click to show/hide)

White kd1 rd8f8 sg8 pc4e4f2h2 Black ke6 bg3 pb2b4e5h4

h#2,5           4 solutions         (8+6)
Anti-Andernach


22 Responses to No.612,612.1 (KS&LP)

  1. Laco Packa says:

    Position may be simpler:
    White Rc7 Pb5 Pe5 Pd4 Pg4 Pa3 Pf3 Kc2 Ph2
    Black Pa6 Ke6 Pd5 Ph5 Pf4 Pc3 Pe3 Pf2
    Stipulation H#2
    Condition AntiAndernachChess
    1.f2-f1=Q=w Qf1-c4=b 2.Qc4-a4=w Qa4*a6 #
    1.f2-f1=S=w Sf1-g3=b 2.Sg3-e2=w Se2*f4 #
    1.f2-f1=B=w Bf1-g2=b 2.Bg2-h3=w g4*h5 #
    C+ by Popeye v4.69

    Probably can also do a complete AUW, I’ll try it tomorrow – today it’s too late:-)

    • Laco Packa says:

      Even before bedtime I managed also AUW:
      White Rb7 Pe5 Pg4 Pd3 Pf3 Kc2 Ph2 Ba1
      Black Pc6 Ke6 Pd5 Ph5 Pf4 Pe3 Pf2
      Stipulation H#2
      Condition AntiAndernachChess
      C+
      1.f2-f1=Q=w Qf1-e1=b 2.Qe1-c3=w Qc3*c6 #
      1.f2-f1=S=w Sf1-g3=b 2.Sg3-e2=w Se2*f4 #
      1.f2-f1=R=w Rf1-b1=b 2.Rb1-b6=w Rb6*c6 #
      1.f2-f1=B=w Bf1-g2=b 2.Bg2-h3=w g4*h5 #

      • Nikola Predrag says:

        Repeating the same mate (Q/Rxc6) should be avoided to show a true AUW, it is surely possible in various ways.

        • Laco Packa says:

          Of course Nikola, you’re right. But it was only the first shot, just to be shown that it is possible AUW. One of the possible ways to remedy shortcomings here:
          White Sg8 Rd5 Bh5 Pc4 Pc3 Pf3 Pa2 Kc2 Pg2 Ph2
          Black Pc6 Ke6 Pf6 Pf5 Pf4 Pe3 Pf2
          Stipulation H#2
          Condition AntiAndernachChess
          C+
          1.f2-f1=Q=w Qf1-e1=b 2.Qe1-h4=w Qh4*f6 #
          1.f2-f1=S=w Sf1-g3=b 2.Sg3-e2=w Se2*f4 #
          1.f2-f1=R=w Rf1-b1=b 2.Rb1-b6=w Rb6*c6 #
          1.f2-f1=B=w Bf1-e2=b 2.Be2-d3=w Bd3*f5 #

          • Kenneth Solja says:

            I like this problem. Mine was only 5 minute achievement where rook-promotion would be the same as queen, so I dropped it away.
            Nice AUW, Ladislav

  2. Kenneth Solja says:

    I manage to do myself the AUW as well:
    White Kc2 Rb7 Pa2 Pd3 Pd4 Pe5 Pg4 Ph2
    Black Ke6 Pc3 Pc6 Pd5 Pf2 Pf3 Pf4 Pf6 Ph4 Ph5
    Stipulation H#2
    Condition AntiAndernachChess
    C+
    1.f2-f1=Q=w Qf1*f3 2.h5*g4 Qf3*g4 #
    1.f2-f1=S=w Sf1-g3=b 2.Sg3-e2=w Se2*f4 #
    1.f2-f1=R=w Rf1-b1=b 2.Rb1-b6=w Rb6*c6 #
    1.f2-f1=B=w Bf1-g2=b 2.Bg2-h3=w g4*h5 #
    I like the knight-promotion here because knight move to e3 is not blocked by pawn, but 1.-Se3 is a check.
    And I dislike bishop-promotion, because it doesn’t end to bishop move, but battery mate.

    • Laco Packa says:

      I offer this correct version, nothing more economical I could not figure out:
      White Sf8 Pf6 Rc5 Pb4 Pg4 Pe3 Pg3 Kb2 Pf2
      Black Pb6 Kd6 Pe6 Pe5 Pg5 Pe4 Pd3 Pe2
      Stipulation H#2
      Condition AntiAndernachChess
      1.e2-e1=Q=w Qe1-d1=b 2.Qd1-b3=w Qb3*e6 #
      1.e2-e1=S=w Se1-f3=b 2.Sf3-d2=w Sd2*e4 #
      1.e2-e1=R=w Re1-a1=b 2.Ra1-a6=w Ra6*b6 #
      1.e2-e1=B=w Be1-d2=b 2.Bd2-c3=w Bc3*e5 #
      Interestingly, despite the external similarity with 578 the moves have other motivation.

      • Nikola Predrag says:

        It’s not easy to say what is the best possible economy, since 4 extra bPs are there only to be captured. The “geometric” potential of the board should be explored to reduce the merely obstructing pieces.

        White Rg7 Pg5 Pe4 Pa2 Ph2 Kd1
        Black Pc7 Pd7 Pd6 Ke6 Ph6 Pe5 Pf5 Pb4 Pb2
        Stipulation H#2
        Condition AntiAndernachChess
        1.b1=Q=w Qb2=b 2.Qf2=w Qxf5#
        1.b1=S=w Sa3=b 2.Sb5=w Sxc7#
        1.b1=R=w Rb3=b 2.Rh3=w Rxh6#
        1.b1=B=w Bd3=b 2.Bb5=w Bxd7#

        Still, some Pawns might be spared in some better scheme.

        • Laco Packa says:

          Well, I did not write anywhere that I managed to make the most economical position … You saved some material – nice work. But it interested me another option to enrich the content. The intention is probably clear from the example that although is correct, but to perfection still a lot lacks.

          White Be8 Be7 Pc6 Pd6 Pc5 Pg5 Pe4 Pf2 Ph2 Kd1
          Black Ke6 Pg6 Pe5 Pb4 Pd4 Bg3 Pb2
          Stipulation H#2.5
          Condition AntiAndernachChess
          1…h2-h3=b 2.b2-b1=B=w Bb1-d3=b 3.Bd3-f1=w Bf1*h3 #
          1…c6-c7=b 2.b2-b1=S=w Sb1-a3=b 3.Sa3-b5=w Sb5*c7 #
          1…d6-d7=b 2.b2-b1=Q=w Qb1-a2=b 3.Qa2-a7=w Qa7*d7#
          1…Be7-f6=b 2.b2-b1=R=w Rb1-b3=b 3.Rb3-f3=w Rf3*f6#

  3. Nikola Predrag says:

    Yes, it was clear that you did not say that your position was most economical.

    I just wished to say that it is hard for composer to be sure about the economy as long as there is one technical piece.
    And it’s not easy to say: “I don’t see how to do it better.”
    But next day some idea might come and third day comes another idea….

    Now, your new idea indeed enriches the content, so go on.
    So far, the biggest flaws are the idle wBe7/wBe8 in the mates by Q/R. But the idea is worth working on it.

  4. Kenneth Solja says:

    White Kc2 Rb7 Pa2 Pd3 Pd4 Pe5 Pg4 Ph2
    Black Ke6 Pc3 Pc6 Pd5 Pf2 Pf3 Pf4 Ph5
    Stipulation H#2
    Condition AntiAndernachChess
    C+
    1.f2-f1=Q=w Qf1*f3 2.h5*g4 Qf3*g4 #
    1.f2-f1=S=w Sf1-g3=b 2.Sg3-e2=w Se2*f4 #
    1.f2-f1=R=w Rf1-b1=b 2.Rb1-b6=w Rb6*c6 #
    1.f2-f1=B=w Bf1-g2=b 2.Bg2-h3=w g4*h5 #
    I got two black pawns off, but solutions stay the same.

  5. Kenneth Solja says:

    I made following version from Laco’s H#2,5
    White Kd1 Se8 Pc4 Pc6 Pd6 Pe4 Pf2 Pg5 Ph2
    Black Ke6 Lg3 Sf7 Pb2 Pb4 Pd4 Pe5
    1…h2-h3=b 2.b2-b1=B=w Bb1-d3=b 3.Bd3-f1=w Bf1*h3#
    1…c4-c5=b 2.b2-b1=S=w Sb1-d2=b 3.Sd2-b3=w Sb3*c5#
    1…d6-d7=b 2.b2-b1=Q=w Qb1-a2=b 3.Qa2-a7=w Qa7*d7#
    1…Se8-f6=b 2.b2-b1=R=w Rb1-b3=b 3.Rb3-f3=w Rf3*f6#
    I manage to save only one piece compared to Laco’s original. Comments?

    • Laco Packa says:

      In that my original problem (and also in your) economy is not perfect. I think I found a little bit better position.
      White Rd8 Rf8 Sg8 Pc4 Pe4 Pf2 Ph2 Kd1
      Black Ke6 Pa5 Pe5 Pb4 Pd4 Ph4 Bg3 Pb2
      Stipulation H#2.5
      Condition AntiAndernachChess
      We may publish it as a joint composition?

  6. Nikola Predrag says:

    That’s very nice.

  7. Nikola Predrag says:

    The scheme is now working very fine and it seems that nothing could be changed. Nevertheless, some other positions of bK could be examined.
    A very nice idea deserves to be presented in a best possible way.

    I hope that before the publication, you will at least improve the obvious details in economy.

  8. Nikola Predrag says:

    My post was in vain. Please, could someone be so nice to explain me the surprising purpose of bPa5/d4?

    • Laco Packa says:

      Simple question, simple answer.
      The composition de jure does not exist until it is published. Its notation was yet in the discussion freely accessible to everyone and anyone can publish it under his own name. But for its possible repair is sufficient time (which currently I do not have) until the award of the tournament.
      I did not want to write it directly to the discussion.
      Therefore, it was (one of several) reasons why I asked Julia to your email address …

      • Kostas Prentos says:

        Laco, the way I understand the matters of publication, priority, etc, according to the Codex, but with an attempt to modernize and include the proposals that appear in a website like this, is the following:

        Any position that is shown, in diagram or notation, as part of the comments to the original problem, is or should be considered a separate problem. Since the formal type of the publication is not followed, regarding names of composers, source, etc., I believe that a good solution would be to consider each proposal in the comments as a version of the original: e.g., Kenneth Solja, (version by Ladislav Packa, or Nikola Predrag). If the involved composers agree on a different arrangement (like in the case of 612.1), they can publish one of the versions as a joint problem.

        Any comment that includes a problem, version, sketch, scheme or anything else, from the moment it appears in this (or any other) website, it becomes public property, meaning that everyone can see it. For example, if I wanted to write an article about AUW in fairy chess, I could use any of the positions appearing in the comments of this problem, and for lack of a better system, I would use the form I proposed earlier (KS, version by X, Julia’s Fairies and the date). As a matter of fact, I believe that all the positions in the comments should participate in the informal tourney and be considered by the judge (sorry Kjell, and future judges), whether they were deliberately composed to participate or they were composed just for fun, or to show an idea, or make a point. Of course, most of the times, these intermediate positions may not have any particular value and only the final version becomes a “diagram”, but it is not unlikely that one of the versions that was “rejected” by the participating composers, may be more appealing to the judge.

        In any case, the positions “shown” here at the comments, take priority as if they were published problems, and therefore any attempt to “steal authorship” would inevitably fail as anticipated.

        • Laco Packa says:

          I agree with you, Kostas, your arguments are clear, logical and above all ethical.
          However, I think that it is better to avoid conflicts and not give them even the slightest chance to be born.

    • JuliaJulia says:

      Pawns bPa5/d4 are moved from No.612.1 by request of Laco Packa, after comment by Nikola Predrag.

  9. Nikola Predrag says:

    OK Laco, thanks for the answer.
    Sometimes I forget that it’s the public place.
    And it would be quite a corrupted mind which would try to steal the authorship in such an obvious way.
    But perhaps you’re right, there are various kinds of minds.

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