No.249 (JV)

No.249 
Julia Vysotska (Latvia)

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Original Problems, Julia’s Fairies – 2013 (I): January – April

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No.249 by Julia Vysotska – A special neutral Zilahi combined with a pin of  mating pieces. (JV)


Definitions:

Nightrider(N): A Rider along a straight line on squares lying a Knight`s move away from each other.

Locust (L): moves on Q-lines but only by capturing an enemy unit, arriving on the square immediately beyond that unit, which must be vacant.


No.249 Julia Vysotska
Latvia
original-08.02.2013
 
249-h#2,5-jv
h#2,5         b) +black Pd4     (1+3+3n)
Nightrider h1
Locust d7
 
 
Solution: (click to show/hide)
 

30 Responses to No.249 (JV)

  1. Petko A.PetkovPetko A.Petkov says:

    No.249 Julia Vysotska
    A very surprising interpretation of Zilahi-theme with neutral pieces! Excellent pin-mates! In this manner it is possible to compose many other interesting problems with similar thematic elements! .

  2. Seetharamanseetharaman says:

    Nicely matched solutions. Excellent demonstration of pinning by Locust! Great !

  3. Nikola Predrag says:

    If this reciprocal mechanism (sacrifice + selfpin of the mating neutral piece on a Locust-line) is indeed original, I will remember it as “Julia’s mechanism”. A wonderful and widely usable idea!

    Still, I would prefer a most economical realization with model mates, without Pawns. For the thematic mechanism, the extension to 2,5 moves is not essential. h#2 with nRb6 and nBe4 would be thematically sufficient..

    6 pieces in total would give 2 model-mate-solutions in h#2 if nQ could be used instead of nB.

    • JuliaJulia says:

      I had an h#2 version, but also with twins and 7 pieces:
      100165.png
      h#2; b) bPe4; 1+3+3n
      Locust e7
      a) 1.Qc1-c4 nBf4-e5 + 2.nLe7*e5-e4 nRc6*c4 #
      b) 1.Qc1-e3 nRc6-d6 + 2.nLe7*d6-c5 nBf4*e3 #
      Now I have to think what your change of nB to nQ can give….
      Thanks!

  4. Georgy EvseevGeorgy Evseev says:

    I tried to remove things which I do not like in this problem.

    The result is:

    White Kh8
    Black Kd3 Qa2 Sd6
    Neutral Qg2 LOa4
    h#2

    I. 1.Sc4 LOnxc4-d4+ 2.Qad2 Qnxd2#
    II. 1.Qb3 LOnxb3-c2+ 2.Se4 Qnxe4#

    The loss of Zilahi was not intended)), but I was not able to keep it when removing bpc3 (absolutely not needed in a).

    • JuliaJulia says:

      Thanks!
      I think this is a bit different kind of this idea. Of course, more economical. But the same time my version seems to me thematically more complicated, and I regret about losing Zilahi in your version (for just 1 pawn?).

      • Georgy EvseevGeorgy Evseev says:

        The problem with this pawn is that is has no function at all. In a sense it is a “hidden” zero-position or double change twin.

    • Nikola Predrag says:

      Georgy, several years ago I made a cycle of black sacrifices with your scheme, but since I had already published a similar miniature without neutral pieces, I didn’t think that the version with nQ should be published. Admittedly, I tried to use nQ to prevent cooks, without noticing the pin 🙁

      White Kg8 Lh5; Black Nd7 Bf6 Kc4 Sf4; Neutral Qh3; h#2, 3.1…
      1.Sd5 Lxd5-c5+ 2.Bc3 nQxc3#
      1.Be5 Lxe5-d5+ 2.Nb3 nQxb3#
      1.Nc5 Lxc5-b5+ 2.Sd3 nQxd3#

      Zilahi can be preserved in Julia’s problem by using bN. I did not try with some orthodox piece.

  5. Petko A.PetkovPetko A.Petkov says:

    About Evseev`s version
    White Kh8
    Black Kd3 Qa2 Sd6
    Neutral Qg2 LOa4
    h#2
    I. 1.Sc4 LOnxc4-d4+ 2.Qad2 Qnxd2#
    II. 1.Qb3 LOnxb3-c2+ 2.Se4 Qnxe4#
    ———————————–
    I think, tha this version has not sufficient value without one of the the main thematic elements in Julia`s problem – the neutral Zilahi! Also Evseev`s version is not full economical: it is possible to change the black Q with the black Rook, for example:
    White:Ka1(1); Black: Kd3, Rb2, Sd6 (3); Neutral: Qg2, LOa4(2)- h#2 2 sol.
    1.Sc4 LOnxc4-d4+ 2.Rd2 Qnxd2#
    1.Rb3 LOnxb3-c2+ 2.Se4 Qnxe4#

  6. Paul Raican says:

    Here is a version without Night-rider:
    White Ka6 Pe4
    Black Pf7 Se5 Bf5 Kf4
    Neutral Bc7 Rg7 Le1
    Stipulation H#3, 2 solutions
    100163.png
    1.Bf5*e4 nRg7-g2 2.f7-f5 nRg2-f2 + 3.nLe1*f2-g3 nBc7*e5 #
    1.nBc7-b6 e4*f5 2.Se5-g4 nBb6-e3 + 3.nLe1*e3-e4 nRg7*g4 #

    • Nikola Predrag says:

      I was thinking about something in this direction:
      White Kg8; Black Kc4 Nh1; Neutral Rb7 Qh3 Lb2; h#2
      100167.png
      1.Nb4 nQc3+ 2.nLxc3-d4 nRxb4#
      1.Nd3 nRb4+ 2.nLxb4-b5 nQxd3#

      • JuliaJulia says:

        Nikola, about your version: yesterday evening I’ve tried to get something the same, I understood your idea, but couldn’t avoid some cooks. The main weakness I see in this your version is the move nRb4 which repeats in the both solutions. Maybe it is possible to make it different, without this repeating, but I can’t for now…

        The second thing: either economy or complexity? – I believe it is a matter of taste, and it might be different in each case. I’ve added one more half-move of nB/nR to make a problem more interesting, to have a longer moves of neutrals, to make it more difficult for solving… Of course, the final is the same in case of 2 or 2,5 moves. As I’ve shown above – I’ve started from h#2. But have decided that h#2,5 is more interesting. Maybe I’m not right to your opinion, but I like 2,5-moves version! 🙂
        Thank you!!

    • JuliaJulia says:

      Yes, Paul, no twins here, but one more half-move is non-thematic (I’d tell more mechanical) here and also makes two solutions not so matching.
      I don’t think it is an improvement to my problem, but I like to see different views and versions. Thank you!

  7. trittenPierre Tritten says:

    I searched in the same way as Nikola, no repeated move in this version:
    White : Kg1
    Black : Kd3 Bh3 Ng8
    Neutral : Qb4 Rb2 La4

    h‡2 (1+3+3)

    1.Nd2 Qnc4+ 2.LOn×c4-d4 Rn×d2‡
    1.Ne4 Rnb3+ 2.LOn×b3-c2 Qn×e4‡

    • Nikola Predrag says:

      Yes, your scheme is closest to Julia’s. I started with it yesterday but there was always need for an additional cookstopper, bP was the best choice:
      W: Kh2; B: Na8 Kf5 Ph4; Neutral Qe3 Rg3 Le2; h#2 2.1..

      But the point was to avoid extra material. Without black fairy pieces, there is a possibility to present a cook as the echo-mate (accidentally not model). But the balance of reciprocal function would be spoiled:
      W: Kh2; B: Qg8 Kf5; Neutral Qe3 Rg3 Le2; h#2 3.1..

      • trittenPierre Tritten says:

        No extra material in this one…

        White : Kb1
        Black : Ke3 Na2
        Neutral : Qc4 Rc2 Lb4

        h‡2 (1+2+3)
        RO=Rose
        LO=Locuste

        1.ROf4 Rnc3+ 2.LOn×c3-d2 Qn×f4‡
        1.ROe2 Qnd4+ 2.LOn×d4-e4 Rn×e2‡

  8. trittenPierre Tritten says:

    Without Nightrider:
    White : Kg1
    Black : Kd3 Bh6 Bh1
    Neutral : Qb4 Rb2 La4

    h‡2 (1+3+3)

    1.Be4 Rnb3+ 2.LOn×b3-c2 Qn×e4‡
    1.Bd2 Qnc4+ 2.LOn×c4-d4 Rn×d2‡

  9. Nikola Predrag says:

    Well Pierre, thist last example shows that it is so easy to give up too early. One look at your example with a Rose and it is clear that instead of ROa2 there could be Ng6. I missed that possibility on saturday.

    Julia, the repeated move nRb4 is not a flaw for itself, because it has very different purposes, a sacrifice and a mate. It would actually be great if nQ could do the same. But yes, it doesn’t look very convincing without a “thematic echo” in the other phase. Anyway, I came to that position in some 5 minutes and I did not intend to search for better possibilities, I was just interested whether your mechanism is possible at all without technical pieces. Pierre’s example with a Rose showed that your original setting was actually the closest to it.

    About complexity, I don’t think that extensions add to the complexity. A power of some mechanism is best shown when the play is concentrated in lesser number of moves.
    The extension in your original problem would be more interesting if in the same phase the same neutral piece would be moved once by White and once by Black.

    The solvers’ preferences are various, personally I am eager to find the author’s original idea, while an abundance of non-thematic and non-original stuff may be only repelling for me.
    I like the hard problems if the idea is hidden due to its originality or paradox.

  10. Georgy EvseevGeorgy Evseev says:

    Six pieces twins without unjustified fairy pieces – minimal change to Pierre’s version.

    White : Kb1
    Black : Ke3 Sg1
    Neutral : Qc4 Rc2 LOb4
    100166.png
    b) Sg1->h3

    a) 1.Se2 Qnd4+ 2.LOnxd4-e4 Rnxe2‡
    b) 1.Sf4 Rnc3+ 2.LOnxc3-d2 Qnxf4‡

  11. trittenPierre Tritten says:

    Two points:

    – I thought that twinning by moving a thematic piece was a flaw
    – I used a Rose because a Nightrider on g6 plays a Knight move in one solution

    Am I right, am I wrong?

    • Nikola Predrag says:

      You are not wrong but also not absolutely right.

      Moving a thematic piece for a twin shold be avoided if possible but we must consider what’s lost, what’s gained. Georgy could have placed bSg1 + bSh3. Each Knight would be half-employed and half-weasel. There would be no twinning by moving a thematic piece. What is better, 2 halfemployed bS’s or one bS twice thematic, but moved for twinning? The tastes may be various, but these two options look the same to me in their essence. Each one of them looks as an artificial attempt to “improve” the other one.

      Since it is a fairy problem with Neutrals and Locust,, one more fairy piece does not look too strange in the position. It plays the thematical 1st black moves but still these moves are not fairy moves, that is probably what Georgy calls unjustified fairy piece. I agree with that, in principle.

      Would a Rose be less unjustified than a Nightrider, that is an irrelevant question. Nightrider is better known and simplier for describing and it’s easy to avoid that “Knight move”, if necessary.

      In principle, it is more important to present an optimal mechanism than the overall economy. Additional cookstoppers might be more acceptable than a lame mechanism.
      3 neutral and 1 black piece make the mechanism. These neutrals are fairy pieces essential for the idea and mechanism, but a black fairy piece is not essential.
      The optimal mechanism with one cookstopper might look as:
      White Kb1;Black Ba5 Ke3 Qh2;Neutral Lb4 Qc4 Rc2

  12. trittenPierre Tritten says:

    Thanks, Nikola
    It’ getting now harder to define what’s the best compromise.
    I tend to prefer the version using two knights.

  13. Peter Harris says:

    I do not know why someone did not suggest the following: in Evseev’s Sg1>h3 version instead of a bS have a bPe2 with twin e2>f4 and a stipulation h#1.5. This would be the logical conclusion to it all – very economical in Time.

    Having seen the two final positions of Julia’s problem commentators say: how can this “mechanism” be shown “most economically” – as though the mechanism was an end in itself. Julia USED the mechanism to create a PROBLEM. The mechanism was not staring one in the face. It had to be found. The enjoyment of the problem lies not only in seeing the mechanism but also in discovering its existence.

    This enjoyment – the pleasurable surprise – will not be experienced by people who do not solve but who simply Click away.

    Concerning the question raised in comments of economy and complexity: of course economy is important. What is also important is that there is difficulty [complexity] – otherwise a problem is not a problem nor is it interesting.

    Allied to the foregoing: the two first moves of the problem, 1. …..nBe4 and 1. ….nRb6 are wonderful introductions to what follows and enrich the problem – very much.

    [The bPc3 about which much fuss has been made is a mere quibble].

    So: commentators have yet to produce a better version than the original.

    • JuliaJulia says:

      Thank you, Peter!! This is exactly what I believed composing this problem..

    • Nikola Predrag says:

      As a principle, my intentions are not to make “versions”/”improvements” or to become an co-author. I try to analise a problem from various apects and discuss it. This helps me in deeper understanig of chess composition. I learn a lot from these comments, even from my own. I hope that anyone might be intrigued to thinking and reaching his own conclusions.

      I did solve Julia’s problem in a couple of minutes, finding possible mates + 1 second to find the first white move.
      So naturally, I was not impressed with such an introduction as a solver. As a composer, I would welcome 2 moves by nR/nB in if these Neutrals were moved once by White, once by Black each phase. The way it is, this introduction is not a flaw but also not thematically important. The idea and the mechanism are complex and beautiful so I (and of course the others) was curious about some aspects of the realization.

      I must go now, later I might add something about your suggestion about bP’s in h#1.5, actually I did have it in mind.

      • Nikola Predrag says:

        I don’t quite believe that anyone would keep such a long attention to this issue, but I’ll give what I promised.

        bNh1 is an active part of the mechanism presented in original No.249. It interferes on the lines, a)d4-b4/b)b5-d3 to alow Black to play a)3.nLxd5-d4/b)3.nLxc6-b5

        nBh7 and nRb1 have no particular reason to be exactly on their initial squares. Their proper positions in the mechanism are exactly on e4/b6, from where they can function in 2 directions/ways. So, a)1nBe4 and b)nRb6 are actually the unnecessary “corrections” which spoil the potentially perfect mechanism.
        Despite those moves and the twinning, the genuine switch which determines the way/direction in which the mechanism will work is the move by bNh1. That is the beauty I see. Simple 1.nBe4/1.nRb6 is adding mild water to a perfect bouquet of an original wine. But OK, adding water gives more to drink, it’s up to a personal taste 🙂

        h#1.5 with bP on e2 or f4 would lack the mentioned ACTIVE line-interference which is very important for the originality and complexity, as an additional cog in a mechanism.
        “Half-employed” thematic bP’s would be lesser flaw than bS’s but the best what I could find was not good.
        (h#2;W:Kb1/B:Pf5,Ke3,Pg2/N:Lb4,Qc4,Rc2;b)rot.90 counterclokwise)

        • JuliaJulia says:

          I understand what you mean. And as you know not always I’m trying to make a problem longer this way. Just in case of e4/b6 for nB/nR their moves are so short and so predictable! If these moves would be longer, it is most probably I’d left h#2. Anyway, I like to see these two different views, I’ll keep in mind all the opinions I’ve got here for the future problems! And if we ever meet I’d still prefer to have some wine without water! 🙂 Thank you! (I don’t like rotated twins much…)

          • Nikola Predrag says:

            Rotated twins could be interesting in some cases, but certainly not here. It was the “best” (read: least bad) that I could do with Pawns. I mentioned it in the previous post as a bad example, perhaps someone curious might try to do it without such twinning.

            Those introductory moves are not very important (or bad after all), I entirely respect your choice as the author. But I just don’t see them as particularly worth and necessary in the thematic content – this means that h#2 could be equally good or better in case of better economy.

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