# No.490 (JM)

 No.490 János Mikitovics (Венгрия) Оригинальные задачи, Julia’s Fairies – 2014 (I): Январь – Апрель   →Предыдущая ; →Следующая ; →Список 2014(I) Пожалуйста, присылайте свои оригинальные сказочные задачи на: julia@juliasfairies.com

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

Определения:

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 Hungary original-21.01.2014   h#3               b) Se4→d2              (2+3)SwappingKingsBerolina superpawn e3 Berolina pawn e2     Solutions: (click to show/hide)   a) 1.BSe3-c1=S Se4-c5 2.BSe2-f1=R[Kg4↔Kf2] Kg4-f4 3.Sc1-d3[Kf2↔Kf4] + Sc5xd3[Kf4↔Kf2] # b) 1.BSe2-f1=S Kf2-g2 2.BSe3-g1=R[Kg4↔Kg2] Sd2-c4 3.Sf1-e3[Kg2↔Kg4] + Sc4xe3[Kg4↔Kg2] # (C+ by Popeye 4.63) Exchange of promotions, chameleon echo. (Author)

### 4 комментария: No.490 (JM)

1. Georgy 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
h#4
b) Sa2->f1
c) Sa2->e2

a)
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. János 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)
SwappingKings

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. János 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. Balasubramanian пишет:

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