No.1233 (JML)

No.1233
Jean-Marc Loustau
(France)

JF-LOGO-1

Original Fairy problems
JF – 2017(I): January – June


Definitions: (click to show/hide)


No.1233 Jean-Marc Loustau
France

original – 26.06.2017

Solution: (click to show/hide)

white PAa6 Qa7 PAb7 Kb8 PAc1 NAc8 Pd4 VAd8 VAe1 NAe6e7 VAf7 NAg1g3 black Ra2 Pb2 Kb5 VAc2 NAd2 PAe2 Pf3 VAf8 VAg2

#2                                              (14+9)
VAO d8, e1, f7, c2, f8, g2
PAO a6, b7, c1, e2
NAO c8, e6, e7, g1, g3, d2


No.1233.1 Jean-Marc Loustau
France

version of No.1233 – 30.06.2017
Dedicated to Shankar Ram

Solution: (click to show/hide)

White NAe8 Qh8 VAb7 NAc7 VAe7 PAf7 Kh7 NAc6 PAg6 VAh5 Pd4 NAa1 VAc1 PAe1 Black VAb8 Pb5 Kf5 Ph4 NAa3 Pb3 Ph3 VAa2 PAc2 NAd2 VAe2 Pf2

#2                                            (14+12)
VAO b7, e7, h5, c1, b8, a2, e2
PAO f7, g6, e1, c2
NAO e8, c7, c6, a1, a3, d2


6 Responses to No.1233 (JML)

  1. Ladislav Packa says:

    Spectacular work. But the theme (a-AB, b-BC, c-CA –> a-C, b-A, c-B) has been already processed, see YACPDB Id 330467.

  2. Jean-Marc Loustau says:

    Thanks, Ladislav, for the « spectacular »…
    About your remark, I knew that this theme (a-AB, b-BC, c-CA –> a-C, b-A, c-B) had been already achieved, I think even I have seen several examples although I am unable to quote them (I did not say my problem is a 1st achievement), and I even thought it had been named because the form is interesting by itself! (may be “Packa theme” would be appropriate???). Your problem with Juraj is very different from mine, so I think there is no question about any kind of anticipation…

    • Ladislav Packa says:

      :-))
      This was really just a note inspired by the fact that this theme is not at all common – mainly because its processing is difficult. It did not concern the naming of the theme and, of course, anticipation is not at all considered.

  3. shankar ram says:

    In the set play, the moves to e4 remove one flight guard and stop one out of the three possible anti-battery mates.
    The spectacular key simultaneously closes the 3 W flight guarding chinese lines. The same B moves now _enable_ the very same flight guard, allowing W to play the one anti-battery mate which guards the other two flights. The reference to a “Chinese A1” is confusing, though.
    Typical Chinese piece style threat paradox effects, combined with additional tries and cyclic piece correspondences in the arrivals on e4 and b6.
    The triple W “anti-Grimshaw” anti-battery on b6 is reminiscent of Jean-Marc’s 1st pr. problem in JF 2012, though that was with normal pieces and here it’s with Chinese pieces, as also a similar multi flight giving key.
    A slight drawback is that 2.Pa-b6 also opens a second guard to a5 and a4 from the WQa7.
    The “Lacny-like” pattern aBC/bCA/cAB -> aA/bB/cC with cyclic duals in one phase is interesting, but may not pass the Gvozdjak test!

    • Jean-Marc Loustau says:

      Thanks Shankar for this comment.
      First the points of minor importance:
      – The reference to specific Chinese A1 is unclear, I agree; the theme A1 implies 2 lines (the other name of the theme being « double interference »), and in this case the 2 lines are in fact a single line, the 2 lines being somewhat overlapped. I admit this is questionable.
      – Of course the problem would not pass the “Gvozdjak test”, and clearly this problem is not a Cyclone! But the spirit of Cyclone appears sometimes in problems which are not Cyclone!
      Now, what you call a “slight” drawback; this is very nice of you for the “slight”, but to be honest, to my eyes, it’s not a “slight” but a “serious” drawback: the spirit of the matrix is in the relationship between the pieces, and this relation is broken. The truth is that sometimes I compose too quickly (also I wanted to send the problem before the deadline), and I just did not see the point (the other truth is that I become older and older!).
      So I’ll send in some minutes a version to Julia, in order to replacing this one. The setting is unfortunately heavier, but clean; the try 1 NAe4 is lost, but I can live with that, the main content is elsewhere.

      • Jean-Marc Loustau says:

        Just some words about the new version, heavier but clean I hope.
        Of course there is no more the drawback noticed by Shankar (that was the purpose!), but there is also only 1 mate (the threat mate) after any King move. Of course duals after black moves which don’t prevent the threat are not relevant (although it has not been always the case: in the 1st quarter of 20th C, duals were often seen as a defect even on moves which don’t prevent the threat); but when there are flights, King moves are special moves, and it’s normal, not only for a solver, to look at what happens on these moves: to my opinion it’s better when there is only 1 mate, what is the case in this new version; when there are 2 mates (what was the case in the previous version) or more, it’s a very very slight drawback, a very admissible drawback, but a drawback nevertheless.
        Now some words about the construction:
        – The VAh5 is a little bit questionable; its function is to guard f7 after 1… Ke6; a white Bg8 would do the job, and in theory it’s better (an orthodox piece is preferable). But I don’t like this Bg8 which is unable to play, completely incarcerated, and incarcerating also the Queen… So my preference goes to this Vao. (There is also another option, but with 1 piece more)
        – It’s possible to save 1 unit by replacing the Pawns b3 and h3 by a white Pao b3; nevertheless this Pao is quite expensive, quite out-of play (just used as a Pao after 1… Kg4), and moreover 2 mates occur after 1… Kf4; so I clearly prefer the 2 Pawns.
        – Finally, in the previous version there was an interesting try by playing on e4 (now c4), with threats 2 A, B, C#, refuted by 1… f2 (now 1… f2) with a Dombrovskys effect in relationship to the set play (1… b2 2 A, B, C#). This is a nice additional content, but very aside the main content and I think it is not really necessary. Nevertheless, it’s possible to get this try by replacing the Pb5 by a white Vaob5: then 1 VAc4? (2 A, B, C#) But 1… b2! But I think this Vao is really out-of-play, and seems to me too “expensive” for the added value…

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