No.490 (JM)

János Mikitovics (Венгрия)


Оригинальные задачи, Julia’s Fairies – 2014 (I): Январь – Апрель

  →Предыдущая ; →Следующая ; →Список 2014(I)

Пожалуйста, присылайте свои оригинальные сказочные задачи на:

No.490 – János Mikitovics – Приятное хамелеон-эхо!  (JV)


Swapping Kings: If a move exposes the opponent’s King to a check under normal rules, then the two Kings are switched, and check or checkmate is newly evaluated, and of course a move is illegal if after such a switch one’s own King is in check.

Berolina-Pawn(BP): Walk and capture are swapped relative to the orthodox Pawn. The Berolina-Pawn moves without capturing diagonally (possibly two squares if it is on the second row of its side) and captures vertically.

Berolina Super Pawn(BS): It is Berolina-Pawn but its moves and captures are respectively extended to the entire diagonal and the entire column.

No.490 János Mikitovics
490-h#3-jmh#3               b) Se4→d2              (2+3)
Berolina superpawn e3
Berolina pawn e2
Solutions: (click to show/hide)

4 комментария: No.490 (JM)

  1. Georgy EvseevGeorgy Evseev пишет:

    I do not think that the use of fairy pawns is justified here.

    While I have not found the setting I am happy with, I think that the following version is a good matter for thought:

    White: Kd2 Sa2
    Black: Ke4 Pc2f2
    b) Sa2->f1
    c) Sa2->e2

    1.f2-f1=R Kd2-e2 2.c2-c1=S[Ke4Ke2] Sa2-b4 3.Ke2-f2 Ke4-f4 4.Sc1-d3[Kf2Kf4] + Sb4*d3[Kf4Kf2] #

    b) wSa2–>f1
    1.c2-c1=R Kd2-e2 2.Rc1-e1[Ke4Ke2] Sf1-e3 3.f2-f1=S Se3-f5 4.Sf1-g3[Ke2Ke4] + Sf5*g3[Ke4Ke2] #

    c) wSa2–>e2
    1.f2-f1=R Kd2-c3 2.Rf1-a1 Kc3-b3 3.Ra1-a3[Ke4Kb3] Ke4-d3 4.c2-c1=S[Kb3Kd3] + Se2*c1[Kd3Kb3] #

  2. Janos MikitovicsJános Mikitovics пишет:

    Yes, you’re right Georgy, but the play differs from the original thought because of the repetition and sequence of promotions.
    Please look at this version and try to develop it as a co-author if you so think:
    White Sa3 Kd3
    Black Kb4 Pb2 Pd2
    h# 4 (2 solutions)

    I.) 1.b2-b1=S Kd3-c2 2.d2-d1=R Sa3-b5 3.Rd1-c1[Kb4Kc2] Kb4-c4 4.Sb1-a3[Kc2Kc4] + Sb5*a3[Kc4Kc2] #

    II.) 1.b2-b1=R Kd3-e3 2.Kb4-c3 Sa3-c4 3.Rb1-b3 Sc4-b2 4.d2-d1=S[Kc3Ke3] + Sb2*d1[Ke3Kc3] #

  3. Janos MikitovicsJános Mikitovics пишет:

    Georgy, of course, the version h#4/2x can be our joint problem, if you like it. I would be happy if you would accept my offer!

  4. S.K.BalasubramanianS. K. Balasubramanian пишет:

    A very fine problem with echo mates. The two solution version is really much better.

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