No.490 
János Mikitovics (Hungary)

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Original Problems, Julia’s Fairies – 2014 (I): January – April

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No.490 by János Mikitovics – A nice chameleon-echo!  (JV)


Definitions:

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
 
490-h#3-jmh#3               b) Se4→d2              (2+3)
SwappingKings
Berolina superpawn e3
Berolina pawn e2
 
 
Solutions: (click to show/hide)

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Georgy Evseev
Georgy Evseev
6 years ago

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] #

János Mikitovics
János Mikitovics
6 years ago

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] #

János Mikitovics
János Mikitovics
6 years ago

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!

S. K. Balasubramanian
S. K. Balasubramanian
6 years ago

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