Original Problems, Julia’s Fairies – 2014 (I): January – April
→Previous ; →Next ; →List 2014(I)
Please send your original fairy problems to: firstname.lastname@example.org
No.490 by János Mikitovics – A nice chameleon-echo! (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
h#3 b) Se4→d2 (2+3)
Berolina 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)
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
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] #
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] #
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] #
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] #
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!
A very fine problem with echo mates. The two solution version is really much better.