Julia's Fairies

No.491 (PH)

Peter Harris (South Africa)


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

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No.491 by Peter Harris – Just an empty board and a fantasy with 7 fairy conditions! (JV)


Double Maximummer: Maximummer for the both, Black and White.

Maximummer (Black Maximummer): Black must play the geometrically longest move or may choose from among longest moves of equal length, distances being measured from the center of each square. Diagonal and oblique distances are measured from the orthogonal coordinates by using Pythagora’s theorem (take the square root of the sum of the squares of the orthogonal distances). All other orthodox chess rules apply.

White Maximummer: Only White must play the geometrically longest moves, Black plays orthodoxal.

Sentinelles PionAdvers: When a piece (not a Pawn) moves, a Pawn of the colour of the opposite side appears on the vacated square if it is not on the first or the last rank, and if there are less than 8 Pawns of that colour on the board.

AntiCirce =Anti-Circe Calvet (the default type): After a capture the capturing piece (Ks included) must immediately be removed to its game array square (necessarily vacant, else the capture is illegal). Captures on the rebirth square are allowed. Game array squares are determined as in Circe.

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.

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 Chess: 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.

Isardam: Any move, including capture of the King, is Isardam illegal if a Madrasi-type paralysis would result from it.

Madrasi: Units, other than Kings, are paralysed when they attack each other. Paralysed units cannot move, capture or give check, their only power being that of causing paralysis.

Lortap: A piece can’t capture if it is controlled.

No.491 Peter Harris
South Africa
491-hs#6-phAdd wK then hs#6                      (0+0)
1 solution
Double Maximummer
Sentinels Pion Advers
Chameleon Chess
Anti-Andernach Chess
Solution: (click to show/hide)

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January 26, 2014 14:00

add wKe2
1.Ke2-d3[+bPe2] e2-e1=S=w 2.Se1-c2=b Sc2-a3=w[+wPc2] 3.Sa3-b5=b[+bPa3] Sb5-c3=w[+wPb5] 4.Sc3-b1=b[+bPc3] Sb1-d2=w 5.Sd2-b1=b[+bPd2] d2-d1=Q=w 6.Qd1-h5=b Qh5*b5[bQb5->d8][+wPh5] #
1.Ke2-d3[+bPe2] e2-e1=S=w 2.Se1-c2=b Sc2-a3=w[+wPc2] 3.Sa3-b5=b[+bPa3] Sb5-c3=w[+wPb5] 4.Sc3-e4=b[+bPc3] Se4-d2=w[+wPe4] 5.Sd2-b1=b[+bPd2] d2-d1=Q=w 6.Qd1-h5=b Qh5*b5[bQb5->d8][+wPh5] #

Dual after Popeye v.4.63

Kjell Widlert
Kjell Widlert
February 3, 2014 00:01

Interesting that the solution is unique without any guiding pieces on the board! It is the ultimate in asymmetry and the ultimate in material economy.

It is not exactly the ultimate in economy of fairy conditions, however… I assume Lortap is just against some cook, for I cannot see a function for it in the solution (7.Kxd4?? is illegal because of Sentinels, so we don’t need Lortap for that).

February 3, 2014 22:22
Reply to  Kjell Widlert

Without Lortap condition, 6.d3-d4=bP would be illegal (self-check). The bPd4 does not give check as long as it is controlled by bQd8.

In the final position, wKc3 cannot move because he would leave a bPc3 (Sentinelles PionAdvers) and that would be illegal according to Isardam condition.

Kjell Widlert
Kjell Widlert
February 3, 2014 23:40
Reply to  Eric

Ah, I forgot that Pd4xKc3 would not leave a bPc3 (which would be Isardam-illegal) as the bP is AntiCirce-transported to c7. Thanks! 🙂

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