Все комментарии

  • From Marcin Kania on Tarasiuk=V - (=0301.11g8h3)

    A great logical idea, but the arrangement of this solution is harmful to its attractiveness.
    It would be good to explain why the white knight goes to c1 through b3 and not d3.
    (5. Nd3? Rd2 6. Nc1 Rd6!! -+ {crucial, surprisingly deep and difficult to notice move that pins the white pawn f6 and forces the white king to leave the good g6 square, 6… Rd1 is waste of time}). After 5. Nb3!! Rb2?? with 6. Nc1 Rb6 is impossible.

    2026/07/16 at 10:58 am
  • From Marcin Kania on Zhukov=A - (+4040.10c6b8)

    After Centurini (or version) and a valuable mutual zugzwang: 5. e8=N !? Bh4 !! (5… Be7 ? 6. Bc7 ! zz-BTM Ba3 7. Nf6 ! Bb4 8. Nd5 Bc5 9. Kxc5 +-) 6. Bc7 Be7 ! = zz-WTM.

    2026/07/15 at 1:32 pm
  • From Marcin Kania on Micu=N - (+0440.10d8e3)

    Calling this study a masterpiece is rather a joke. The introduction (1. c7 Rc1 2. Rg3) and the starting position are very difficult to accept. A try that could possibly justify the attacked: rook on g7 and the bishop on f3 (2. Rf7?) is rejected by 2… Rxc7 and 2… Bf4 and 2… Bg5+.
    I recommend the following version:

    2026/07/14 at 8:55 pm
  • From Marcin Kania on Pervakov=O - (+1034.22d2h4) A

    Is the trivial model mate after 13… Kh5 on the edge of the board so beautiful that it negates the natural 13… Kg4 14. Qg5#?
    In the latter case, the g3-square is additionally attacked by the queen, but through the black king. Mate is economical, and the contested g3-square (and two others!) are attacked by the white king, and it’s impossible to even hypothetically improve this position, as it’s the only way (!) to mate with the KQP/K-set in the center of the board. Thus, this may be the valuable and surprising ending expected here.

    2026/07/11 at 1:12 pm
  • From Marcin Kania on Tomeo=I - (=0034.11d8h5)

    A beautiful, perfect, and emotional classic, more discovered than created, which many unnecessarily consider a weakness and add unnecessary or poorly chosen introduction.
    You can find such emotional works (many versions) in the e-book: «In Search of Beauty» 2025:
    https://www.mediafire.com/file/8qw5rtjf7q4rsm8/InSearchOfBeauty.zip/file

    2026/07/09 at 11:19 am
  • From Marcin Kania on Koranyi=A - (+1643.01g5h8)

    Such a masterpice idea needs a better introduction.

    2026/07/07 at 8:17 pm
  • From Marcin Kania on Troitzky=A - (=0130.11h4f8) U2

    This is more of a mistake than a version: 1… Kg7 2. Kh3 and 2. Rf3
    Therefore, the white king must stand on g4 (original).

    2026/07/05 at 12:26 pm
  • From Marcin Kania on Hoch=Y - (+0404.30h1f1) (v) A

    A masterpiece! The original author (version or based on) has at least 50% credit. Here, the other full 50% goes to Serhiy. A wonderful, perfectly created introduction!

    2026/07/04 at 8:32 pm
  • From Marcin Kania on Rusinek=J - (=0036.40c8e7) A

    Do you respect Kasparyan? His commentary in the book differs from the one here. There was added the crucial word «stalemate» at the end of the commentary, along with the move 9… Ne4. It changes the image of this study.

    2026/07/03 at 8:22 pm
    • From Didukh on Rusinek=J - (=0036.40c8e7) A

      Rusinek in his book ends the solution of his study with «9…Ne4 pat z zamurowanym dorobionym gońcem i związaną dorobioną wieżą». Kasparyan ends it with «9.Rb7 and inevitable stalemate» which is not true because Black can avoid it, for example by taking the rook.

      2026/07/04 at 12:36 am
      • From Marcin Kania on Rusinek=J - (=0036.40c8e7) A

        It seems Kasparyan didn’t like the stalemate-ending move 9… Ne4. Why not 9… Ke7 (dual ?)?
        9… Ne4 would make sense if there were a logical try 5… Bf1? 6. b8=Q Ba6+ 7. Qb7 8. Ne4 8. Qxa6 Nd6+ 9. Qxd6
        But then you can also 6. b8=N, and unfortunately, the thematic 6. b8=R, which shows that 5… Ne5 is just a simple knight repositioning (to avoid duals) without any depth. This was reported by Marek Kwiatkowski on Problemista.eu.

        2026/07/04 at 12:03 pm
  • From Didukh on Kasparyan=G - (+0044.13d3b6) (c) A

    Картина «Художники».

    2026/07/01 at 10:26 pm
  • From Marcin Kania on Kasparyan=G - (+0044.13d3b6) (c) A

    This is a fantastic realization of a deep idea by the immortal Kasparyan, with an incredibly difficult to imagine here and humorous ending: the black king’s fork attack is met by the white king’s fork defense. A masterpiece of logic!

    2026/07/01 at 7:45 pm
  • From Marcin Kania on Platov=V Platov=M - (+0011.23g3e3) A

    The correct White’s move is 6. Bf4 instead of 6. Bg5 (main line);
    6. Bg5 (or 6. Bh6) is a waste of time, what Black proves:
    6. Bg5 d3 7. Kg4 Kc3! 8. Kf5 d2 9. Bxd2 Kxd2 10. Kg5 Ke3 11. Kh6 Kf4 12. Kxh7 Kg5 13. h6 Kf6 14. Kg8
    After 6. Bf4! d3 7. Kg4 (Kh4 or) d2 (now move 7… Kc3 has no sense) 8. Bxd2 and as above but one tempo faster.
    And now it is wonderful work with a good logical try 2. Nf3?

    2026/06/29 at 2:58 pm
  • From Didukh on Korolkov=V - (=0434.11h4a2) (v) A

    Теперь играют все фигуры.

    2026/06/27 at 3:42 pm
  • From Stavrietsky on Stavrietsky=A - (+0040.75h2f2) U2

    А что будет, если добавить белую пешку на е5? Дуали не будет.Но будут ли другие проблемы?

    2026/06/24 at 10:22 pm
  • From Jan Sprenger on Stavrietsky=A - (+0040.75h2f2) U2

    Harold van der Heijden once wrote an article «A minor dual is not a big deal». Here we have a major dual, but still I am not sure how much of a problem it is. Of course, strictly speaking the study must get 0 points. But the dual is sufficiently hidden, and the solution sufficiently obvious that the aesthetic impression of the systematic movement survives.

    2026/06/16 at 5:46 pm
  • From Jan Sprenger on Stavrietsky=A - (+0040.75h2f2) U2

    I guess the main line should go 14. Kd1 (zugzwang) 14… Ka1 15. Kc1 and 1-0.

    I am not 100% sure about the alternatives on White’s 12th move. Is 12. Kf4 Kb2 sufficient to keep the balance? (White’s bishop cannot move because of Kxa3. If 12… Bd3 then 13. Be8 and 14. Bc6.) For example 13. Ke5 Bb1 14. Kd4 Bc2 15. Be8 and now Kxa3 can be countered with Kc3. Same question for 12. Be8.

    However, continuing the main plan with Ke2, Kd2 and Kd1 is clearly the safest and fastest path to victory, so if there are duals, perhaps they are not a big deal.

    2026/06/15 at 5:16 pm
    • From Jan Sprenger on Stavrietsky=A - (+0040.75h2f2) U2

      Serhiy: if we are in the business of giving duals, I think the following line is clearer: 12. Kf4 ~ 13. Ke5 ~ and now either 14. Kd4 followed by Be8 and Bc6 (answer Kxa3 with Kc3 and 1-0) or 14. Be8 if the a-pawn does not hang. I see nothing Black can do about this.

      Of course your line is more forced, and this is a good thing, but it is very long. You even have to realize that at the end, there is no accidental perpetual or h5 falls with check.

      2026/06/15 at 11:32 pm
  • From Steffen Nielsen on Stavrietsky=A - (+0040.75h2f2) U2

    Probably my mistake (adding this line). I will make sure to correct it in the final award.

    2026/06/15 at 3:58 pm
  • From Stavrietsky on Stavrietsky=A - (+0040.75h2f2) U2

    14.Bxb1. А будет ли здесь выигрыш? Вариант в скобках.14.Be4. А не лучше ли взять черного слона. Но зачем 14.Be4, Чтобы слона побыли? Хоть слоном, хоть королем,

    2026/06/15 at 2:59 pm
  • From Didukh on Bazlov=Y - (+3131.30c7g4) (v)

    Лучше оригинала, потому что начальная позиция проще — по две фигуры у сторон, вступление элегантнее и понятнее, ферзи оказываются под ударами без грубых взятий. Шедевр.

    2026/06/12 at 9:55 pm
  • From Didukh on Platov=V Platov=M - (+0011.23g3e3) A

    Этюд напоминает обезьяну, которая может попасть в категорию красивых, благодаря вариантам 3…Qa5 4.Bxd4! и 3…h6 4.Be5 Qc3 6.Bf4# или в категорию умных — после скучного 3…Qxc1 выигрывается темп.

    Красоту и ум можно совместить в одном варианте:

    2026/06/12 at 8:18 pm