No.253 (PH)

 No.253 Peter Harris (South Africa) Original Problems, Julia’s Fairies – 2013 (I): January – April →Previous ; →Next ; →List 2013(I) Please send your original fairy problems to: julia@juliasfairies.com

No.253 by Peter Harris – A nice four-men which is difficult for solve! (JV)

Definitions:

Sentinelles: When a piece (Pawn excluded) leaves a square outside the first and last rows, it leaves a Pawn of the color of the side that played unless 8 Pawns in this color are already on the board.

Super-Circe: When captured, a piece is reborn on any free field on the chess board without causing self-check or selfmate. Possible is also removal of captured piece from the board. The Pawns (white, black, neutrals, half- neutrals) can be reborn on the first or eight row also. When reborn on the first row (for Black) or on the eight row (for White) the promotion is obligatory. When reborn on the first row (for White) or on the eight row (for Black) the Pawns are immovable.

Chameleon Chess: All pieces on the board which are displayed as orthodox  Q, R, B, S,  are Chameleons. A Pawn can promote only in Chameleon-pieces.

Chameleon: On completing a move, a Chameleon (from classical standard type) changes into another piece, in the sequence Q-S-B-R-Q…  Promotion may be to a chameleon at any stage in the cycle.

Anti-Andernach: A piece (excluding King) changes its color after any non-capturing move. After capture, the piece retains its color. Rooks on a1, h1, a8 and h8 can be used for castling, provided the usual other rules for that move are satisfied. After castling, Rooks do not change color, If White makes a non-capturing move with neutral or halfneutral piece, that piece becomes black and vice versa.

 No.253 Peter Harris South Africa original-14.02.2013 Dedicated to Petko A. Petkov   ser-hs#6    b) wBa7→wRa7        (2+2)SentinellesSuper-CirceChameleon ChessAnti-Andernach Chess     Solution: (click to show/hide)   a) 1.Kd7-c8[+bPd7] 2.d7-d6=wP 3.Bc7xd6=R [+wPa6][+bPc7] 4.Rd6xa6=Q [+wPb8=wS][+bPd6] 5.c7-c6=wP Sb8xa6=B [+bQd8] + 6.Kc8-c7 # b) 1.Kd7-c6[+bPd7] 2.Kc6-c5[+bPc6] 3.Bc7-b6=wR[+bPc7] 4.Kc5xb6 [+wRd4][+bPc5] 5.c5xd4 [+wRc4] Rc4xd4=Q [+bPa1=bR][+wPc4] + 6.Ra1xa7=Q [+wRc5] #

5 Responses to No.253 (PH)

1. seetharaman says:

Including the stipulation, five fairy conditions! Perhaps a record ?

• Peter Harris says:

I am so pleased you counted the number of conditions.

I wonder whether you solved it.

But it is not a record. Five years ago in feenschach there was the following problem of mine:

EMPTY BOARD

stipulation: add a piece then h#3.5

conditions:

maxi
antiandernach
anticirce
einstein
republicanchess.

I may have done better than this – at some time or another.

• Julia says:

Peter, very interesting about that empty board problem!! May I ask you to show a solution? And have you tested this your solution and how? With popeye? Thanks!

• Peter Harris says:

Here is the solution to the empty board problem:

+ wQa7 then:

1…Qa7-e7=bR[+bPa7] 2.Re7-e1=wB[+wPe7] Be1-a5=bS 3.Sa5-b7=wP[+wPa5] b7-b8=bP 4.b8-b5=wP e7-e8=bP[+bKa6]# Popeye.

All Ps are necessary for #.

• seetharaman says:

I did not try to solve it. I am not a good solver and the many fairy conditions makes it a little intimidating to try ! Generally, I like to see the problems with solutions and comments of author/others.