vnefyodov-smallNo.924 
Vladislav Nefyodov
(Russia)

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Original Problems, Julia’s Fairies – 2015 (II): July – December

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No.924 by Vladislav Nefyodov – AUW with Cyclic Zilahi. (JV)


Definitions:

Circe: Captured units (not Ks) reappear on their game-array squares, of the same colour in the case of pieces, on the file of capture in the case of pawns, and on the promotion square of the file of capture in the case of fairy pieces. If the rebirth square is occupied the capture is normal.

Anti-Circe (Cheylan): When a piece captures (including King), it must come back to its rebirth square. If this square is occupied, the capture is forbidden. A Pawn capturing on its promotion rank promotes before it is reborn. The captures on the rebirth square are forbidden.


No.924 Vladislav Nefyodov
Russia

original – 10.10.2015
Dedicated to Michael Grushko for his 60

Solutions: (click to show/hide)

White Kc4 Pd5 Pe2 Black Ka3 Pe4 Neutral Rd2 Sd1 Pc2

h#2               4 solutions       (3+2+3)
Circe
Anti-Circe (Cheylan)


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Kenneth Solja
Kenneth Solja
4 years ago

In the solution where pawn promotes to rook, isn’t there a possibility to change the move order:
1. nPc1=nR+ nRb1 2. nSc3 nSxb1(+nRa8)#
or the rook can be captured also in d1:
1. nPc1=nR+ nSc3 2. nRd1 nSxd1(+nRa8)#

Joost de Heer
Joost de Heer
4 years ago
Reply to  Kenneth Solja

In Anticirce type Cheylan, a piece can’t capture on its rebirth square.

Kenneth Solja
Kenneth Solja
4 years ago
Reply to  Joost de Heer

Sorry, I don’t understand this because in solution:
1. nPc1=nR+ nRb1 2. nSc3 nSxb1(+nRa8)#
I have only changed W1 and B2 moves places, all these moves are happening also in author’s solution although knight is capturing to b1 which is it’s rebirth square.

Why then solution:
1. nPc1=nR+ nSc3 2. nRd1 nSxd1(+nRa8)#
isn’t possible?
There is nothing capturing to it’s own rebirth square.

Joost de Heer
Joost de Heer
4 years ago
Reply to  Kenneth Solja

Because there’s a typo in the solution, Rb1 should be Rd1.

Kenneth Solja
Kenneth Solja
4 years ago
Reply to  Joost de Heer

So this is correct solution:
1. nPc1=nR+ nSc3 2. nRd1 nSxd1(+nRa8)#
?

Joost de Heer
Joost de Heer
4 years ago
Reply to  Kenneth Solja

Yes (last moves are 2. nRcd1 nSxd1(nSd1→b1)(+nRa8)# to be precise).

Michael Grushko
Michael Grushko
4 years ago

Thank you,Vladislav!!