JV problem No.22

Julia’s Published Problems

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Thank you for the interest! Your comments are very welcome! – Julia


No.22 – finished 11.01.2012 – This time I have a pair of two Nightriders instead of two Knights I often had before 🙂 . Chinese pieces are also one of my favorite.


No.22
99745 /

Julia Vysotska

Phenix – 2012, No.216, #6495

hs#3       2 solutions         7+9
Leo g2; Pao h1; Vao d8;
Nightrider h4, h3
 

Solutions:

I. 1.Sd4 Nf5 + 2.VAh4 f3 3.LEg4 + Nf4 #

II. 1.Sc5 Nd1 + 2.LEh3 d5 3.VAe7 + Nd6 #

Transformation of initial N-N/PAO battery into two new batteries: LEO-N/PAO and VAO-N/PAO. Opening of lines for the white LEO/VAO with the simultaneous vacation of squares for black pieces (NN), Umnov, change of functions of black pieces (NN) and white pieces (LEO/VAO), creation of masced white anti-batteries, model mates through double checks.

No.22a – Improved version of No.22 based on the comments below, with a gratitude for all advises!
100112 /

Julia Vysotska

Version of No.22

hs#3       2 solutions         7+8
Vao e6; Pao f3, g1;
Nightrider g4, g3
 

Solutions:

I. 1.Sa3-c4 Ng4-e5 + 2.VAe6-g4 e4-e3 3.PAf3-f4 + Ng3-e4 #

II. 1.Sa3-b5 Ng3-c1 + 2.PAf3-g3 c6-c5 3.VAe6-d7 + Ng4-c6 #

Transformation of initial N-N/PAO battery into two new batteries: PAO-N/PAO and VAO-N/PAO. Opening of lines for the white PAO/VAO with the simultaneous vacation of squares for black pieces (NN), Umnov, change of functions of black pieces (NN) and white pieces (PAO/VAO), creation of masced white anti-batteries, model mates through double checks.


28 Responses to JV problem No.22

  1. Paul Raican says:

    In this position, Leo g2 could be replaced by a white Pao g3. And in this case, bPc7 is superfluous. (C+)

  2. JuliaJulia says:

    Yes, Paul, thanks! – you’re right:

    100104.png
    1.Sb3-d4 Nh4-f5 + 2.VAd8-h4 f4-f3 3.PAg3-g4 + Nh3-f4 #
    1.Sb3-c5 Nh3-d1 + 2.PAg3-h3 d6-d5 3.VAd8-e7 + Nh4-d6 #

    I’m thinking now why have I left that Leo? Honestly, don’t remember, it was in January. Use of PAO looks so logical here! Most probably, had another version, than changed it to this one, and didn’t realize, that Leo can be changed.
    I’m afraid, I have many such mistakes.. Either too inattentive or hurrying too much?

    • Dmitri TurevskiDmitri Turevski says:

      Seems also to work: -wRa8, -bPb5, +wQa6

      Pity it can not be setup with dual avoidance effect by wS arrival: wRg3, wBd8 (instead of Pao and Vao) and wSb3->e6:
      1.Se6-d4 Nh3-d1+! (Nh4-f5+?) 2.Rg3-h3 etc. Perhaps this was original intent?

      • SeetharamanSeetharaman says:

        Similar dual avoidance effect is there. Now, the arrival effect of the white knight is used instead of non-arrival !

        • Dmitri TurevskiDmitri Turevski says:

          Pardom me, Seetharaman, but i disagree completely.
          1.Sb3-d4 / 1.Sb3-c5 has 2 positive effects: guarding b3 and creating a hurdle on g4-b4/e7-b4. No negative effects here.
          Dual avoidance is always connected to negative effects (which do not allow duals to work), not present in diagram position, in this case closing of g4-b4/e7-b4 would avoid dualistic checks on these lines.

  3. Paul Raican says:

    The replacement of a Leo with a Pao is just a matter of economy. And the economy is a principle in composition. Now, I have a suggestion to eliminate the Rook a8! (but first I wait your permission)

  4. Paul Raican says:

    Pieces
    White VAc8 Sf8 Kg8 Ph7 Sa3 PAf3 Rb1
    Black Sf7 Pc6 Pa5 Ka4 Pe4 Ng4 Ng3 PAg1
    HS#3 2 sol C+

  5. JuliaJulia says:

    Thanks to all! I have what to think about here..
    So, 3 main versions for now (I like systematization!):

    V.1. by Paul Raican
    100104.png
    hs#3; 2 solutions; (7+8)
    Vao d8; Pao g3, h1; Nightrider h4, h3
    —————————–
    V.2. by Dmitri Turevski
    100109.png
    hs#3; 2 solutions; (7+7)
    Vao d8; Pao g3, h1; Nightrider h4, h3

    I. 1.Sb3-d4 Nh4-f5 + 2.VAd8-h4 f4-f3 3.PAg3-g4 + Nh3-f4 #
    II. 1.Sb3-c5 Nh3-d1 + 2.PAg3-h3 d6-d5 3.VAd8-e7 + Nh4-d6 #
    —————————–

    V.3. by Paul Raican
    100110.png
    hs#3; 2 solutions; (7+8)
    Vao c8; Pao f3, g1; Nightrider g4, g3

    I. 1.Sa3-c4 Ng4-e5 + 2.VAc8-g4 e4-e3 3.PAf3-f4 + Ng3-e4 #
    II.1.Sa3-b5 Ng3-c1 + 2.PAf3-g3 c6-c5 3.VAc8-d7 + Ng4-c6 #

    —————————–

    For sure V.1. is better than mine, and it is my fault, that I haven’t changed Leo to Pao.
    But I’m not sure about V.2 and V.3.
    There’s economy of 1 more piece in V.2, but Q is used. If I had no Rook on c1 guarding c4 I’d believe that usage of Q is justified – it would guard 5 squares.
    Now it guards 4 squares (or needs to) and saves 1 pawn. I don’t know really! Although, if it would give a Meredith, I’d be more sure about using it! 🙂
    V.3 (position shifted to the left) has the same number of pieces as V.1 but there’s Sf7 which guards h8 instead of Pg7 in V.1 which has no functions, just stops cooks.

    Maybe some more tricks?

    Dmitri, you’re right that the initial idea was to use B and R instead of VAO and PAO.
    But.. couldn’t find any idea for the black’s second move, plus more cooks. I still believe it is more complicated to compose it with B&R, but I’d be interested to see any idea if you (or anyone also) have it!

    • Dmitri TurevskiDmitri Turevski says:

      I think the main problem with B&R setting is not cooks or black second move (line clearance by the black pawn works just fine), but that it requires some different matrix. Look:

      Scheme:
      6(r3)1

      (in case i have failed the diagram:
      white Bc8 Kg8 Pf2h7 Sd6f8 PAb2 Rf3
      black Pa5b3c6e4 Sf7 PAg1 Ka4 Ng3g4)

      1.Sd6-c4 Ng3-c1 + 2.Rf3-g3 c6-c5 3.Bc8-d7 + Ng4-c6 #
      works just fine, but it is the only solution shown by Popeye as
      1.Sd6-b5 Ng4-e5+ 2.Bc8-g4 e4-e3 3.Rg3-g4+ Ng3-e4# fails because of 3…Ne5-c4

      • JuliaJulia says:

        Thanks, Dmitri! Yes, in this setting B/R doesn’t work. (PAb2 shouldn’t be here as technical piece, but I understand, this is just an example how the idea could work).
        If it worked, then it would be a question what is better, and I guess that it would be the one with B/R if the same thing can be shown without fairy pieces VAO/PAO.

        Speaking about B/R, I meant some a bit different scheme, where the main theme works – Transformation of initial N-N/PAO battery into two new batteries: LEO-N/PAO and VAO-N/PAO. But the other play or matrix should be different, of course. I haven’t found a good one that time. But I don’t think that it doesn’t exist at all. Most probably it does!

        But this would be either some absolutely different problem or kind of version of my this problem, but most probably not an improvement of it. Or?

        • Dmitri TurevskiDmitri Turevski says:

          I think if it’s ok to replace Pao with Kangaroo then this still can be done pretty close to the original setting:
          3B3(n3)
          hs#3 2.1… C+
          h1: Kangaroo, h2, h3: Nightrider

          1.Sb2-d3 Nh2-g4 + 2.Rg2-h2 d5-d4 3.Bd7-e6 + Nh3-d5#
          1.Sb2-c4 Nh3-f7 + 2.Bd7-h3 f3-f2 3.Rg2-g3 + Nh2-f3#

          • JuliaJulia says:

            Well, you’ve proved that the realization of my scheme is possible with B & R, but I believe you agree that it is too high prize for it – too heavy construction, too many white technical pieces, no model mate in the 2nd solution where Nf7 and Ra8 guard h8….

            But I’m thankful to all of you for analyzing this problem and even more – for “returning me back to it”, as it was composed relatively long ago, and I had added it to the site without taking a look at it again… But today, even being at work 🙂 , had to think about it…
            So, I’ve added an improved version to this problem, using Paul’s advises about changing Leo to Pao and position’s shift to the left,
            plus have moved VAO to more active position,
            and have changed Sf8 to Bf8 (it looks as it guards Sa3 to make it more complicated for solvers 🙂 ) – see No.22a on the top of the page, right after No.22.

            Version of No.22
            100112.png
            hs#3; 2 solutions; (7+8)
            Vao e6; Pao f3, g1;
            Nightrider g4, g3

            1.Sa3-c4 Ng4-e5 + 2.VAe6-g4 e4-e3 3.PAf3-f4 + Ng3-e4 #
            1.Sa3-b5 Ng3-c1 + 2.PAf3-g3 c6-c5 3.VAe6-d7 + Ng4-c6 #

            • Dmitri TurevskiDmitri Turevski says:

              I’m not sure i would so readily agree about the too high price.

              Put it this way: is it ok to use white chinese pieces that do not exhibit anything specific (antibatteries, different move/capture rules etc) just for the sake of light construction?

              • JuliaJulia says:

                Dmitri,
                Sorry, I still consider that your construction is aesthetically not good. I haven’t tried to improve although..

                But about my chinese pieces – I believe I have nice tries here, which doesn’t exist (and can’t exist in this scheme) in your version:
                The first moves can’t be like:
                – 1.VAd7+? c5 – and it’s too early to create an anti-battery with Sb5 and the same time 2. PAg3 is impossible too as g3 is still occupied;
                – 1.PAf4+? e3 – and now Sc4 and 2.VAg4 doesn’t work.

                Of course, each of us has his own view! I just write about mine! Thank you for sharing your thoughts!

                • Dmitri TurevskiDmitri Turevski says:

                  You should not be sorry at all!

                  It’s not just that everyone is entitled to have his own view. I believe the controversy of opinions on aestetics and critical thinking donates greatly to the development of problem chess.

                  It is better to disagree 🙂

                • Dmitri TurevskiDmitri Turevski says:

                  The observation that in Vao/Pao setting there’s a choice of delivering the final check either directly or with antibattery, while in B/R setting battery check is not an option is absolutely valid. Well spotted!

                  A very interesting device and discussion indeed.

  6. Paul Raican says:

    Now, using a white Vao at f8, black Pawn a5 is not necessary.

    • JuliaJulia says:

      Oh, for some time I thought, you’ve written it by mistake.. But have solved your position and understood what you mean.
      The first move is different then in the 2nd solution:
      I. 1.Sa3-c4 Ng4-e5 + 2.VAe6-g4 e4-e3 3.PAf3-f4 + Ng3-e4 #
      II. 1.Rb1-b5 Ng3-c1 + 2.PAf3-g3 c6-c5 3.VAe6-d7 + Ng4-c6 #

      This is the first time here when I don’t agree with you…
      In this case we have a bit unequal solutions:
      after 1.Rb1-b5 in the 2nd sol. the black pawn 2…c6-c5 not only opens line for 3.VAe6-d7 but also includes VAf8 to guard a3. (the move c6-c5 is also more predictable in this case, as Sa3 should be guarded somehow)
      But in the 1st solution we don’t have the same thing.
      And in the 1st solution VAf8 is really just a chinese technical piece…
      But thanks for showing!! And for all this discussion!!

      • Paul Raican says:

        @Julia: Sincerely, I posted this last sentence only to see your opinion (sorry for that). And was a pleasant surprise to see that your opinion is plenty the same with mine! Congratulations, you really has become a problemist, and one very subtle. And this only after two years of composition.
        Now, I have another version: //diagram inserted by Julia, hope you don’t mind!//
        100114.png
        Pieces
        White VAc8 Bf8 Kg8 Ph7 Sa3 PAf3 Rb1
        Black Pf7 Pc6 Pa5 PAh5 Ka4 Pe4 Ng4 Ng3 PAg1
        Stipulation HS#3
        The intention is unchanged, but we have here a try:
        1.Rb1-d1? Ng4-h6 + 2.Kg8-h8 e4-e3 3.Rd1-d4 + Nh6*d4 + 4.Bh6! Deserves this try a chinese technical piece?

        • JuliaJulia says:

          Is this an exam No.2? 🙂
          No, I don’t think that a try is sufficient for using fairy piece as technical piece.
          If only 2 thematic and interesting tries – to have them in the both solutions…. then maybe…

          Another question is what is try in hs# problem?
          What you have shown is more like some solution which doesn’t work, but it is not related to the right solutions.
          In my case a try was a different moves’ order.
          Or (as I understand) a try can be some another move (a wrong one) inside the right solution – like on the 2nd, 3rd etc. move. But something completely different? Probably – not.
          I don’t know where to find all this theory…..

  7. SeetharamanSeetharaman says:

    Thanks Paul Raican, Dmitri Turevski and Julia for an interesting discussion of this problem.

  8. Diyan Kostadinov says:

    Interesting problem and discussion. It looks that Dmitri is right that the basic idea can be realised with white orthodox pieces. One example with Forsberg twins:
    //diagram inserted by Julia, hope you don’t mind!//

    100115.png

    White: Kg8, Rb1,f3, Be6,f8, Ph7
    Black: Ka4, Sf7, Bg4, PAOg1, Nightrider g3, Pa5,c6,e4
    HS#2,5 b) g4=Nightrider (6+8) C+

    a) 1…Bh5+ 2.Bg4 e3 3.Rf4+ Ne4#
    b) 1…Nc1+ 2.Rg3 c5 3.Bd7+ Nc6#

    Of course in case of using white orthodox pieces the first white moves by the Knight are eliminated because the hurdle piece is not needed.

    By the way – Julia, in your comment about thematic complex you noted “creation of masked white anti-batteries”. Actually there are no “anti-batteries”, there are just checks by Chinese pieces after open of lines. In “anti-battery” the check should be given by playing of the hurdle (front) piece on the thematic line of the rear piece.

    • Nikola Predrag says:

      The technical details of this discussion are interesting but generally I feel like an alien from another “galaxy”, where some rather different concepts exist.
      The idea of the replacement of a black hurdle by a white one on a black battery line is very nice. But the great weakness is inbuilt already in the very mechanism for the realization of the idea – the thematic white hurdels have a superfluous power before the final part of a helpselfmate (s#1) -> there could be Pawns on h4/h3 in the final.

      I am still not sure what to think about the combined stipulation h#+s#, but at present I think that in hs#, the “weasels” should be treated more severely than in h#.
      Nevertheless, the economy/necessity of the fairy pieces is an intriguing issue. A couple of days ago I tried with exactly the same position with the twin as proposed by Diyan. The change of the thematic black piece is questionable, it seriously decreases the effect of reciprocal functions of thematic black pieces. This provokes a theoretical question, what is the essence of the “Fosberg” twins?
      2 different thematic pieces (on the same square in 2 twins) is not a very good way of twinning in principle; 3,4 or more could make the Fosberg twins. But it depends on the content of some particular problem.
      The reciprocal functions of Bg4/Ng3 and Ng4/Ng3 in these “Fosberg” twins disappear, actually they never come to existence.

      I don’t see white anti-batteries, especially not the masked ones. Model mates through double checks is a concept beyond my understanding.

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