Fairy chess composition

# No.745 (PH)

 No.745 Peter Harris (South Africa) Original Problems, Julia’s Fairies – 2015 (I): January – June    ?Previous ; ?Next ; ?List 2015(I) Please send your original fairy problems to: julia@juliasfairies.com

No.745 by Peter Harris – An exotic mate with 4 fairy conditions and white royal S! (JV)

Definitions:

Royal piece: Piece that executes a function of the King on the board.

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.

Sentinels Pion Advers: 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.

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.

 No.745 Peter HarrisSouth Africaoriginal – 07.03.2015 Solutions: (click to show/hide) white royal se5 black kh6 qc4 rh2 bd2 ser-#7                                 (1+4)Royal Se5White MaximummerSentinelles Pion AdversSuper-CirceChameleon Chess 1.rSe5*c4=rB [+bQb1][+bPe5] 2.rBc4-g8=rR[+bPc4] 3.rRg8-a8=rQ { } 4.rQa8-f3=rS 5.rSf3*e5=rB [+bPh5][+bPf3] 6.rBe5*h2=rR [+bRh7][+bPe5] { } 7.rRh2*d2=rQ [+bBg6][+bPh2] # { (C+ by Popeye 4.69)}

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Kjell Widlert
March 8, 2015 21:15

The most interesting part is why the SuperCirce rebirths are precise … well, the rebirths on h5/h7/g6 are easy, but the real question is why Qb1 is forced.

The reborn bQ must guard g1 so that 3.rRa8 is legal (white maxi!). But on the lines g1-a7 or g1-g8 she would stop the mate, so she must be on the first row. On a1/h1, she would guard a8, and on other squares like d1 she would stop the mate – so only b1 remains.

peter harris
March 9, 2015 21:02

Good Kjell.

I wonder whether you noticed that the bQb1 does double duty. It enables not only 3. Ra8 but also 6.B*h2. I thought of having a bP on g7. Solvers would then have had to look twice as far ahead [to the penultimate move] to see the need bQb1. But I did not want to block g7 and lose the effect that it need not be blocked.

If g7 was blocked Sentinelles Pionadvers would be hard to justify – and a problem could quite easily be made without it.

I tried to make the problem as simple as possible to encourage solving. Perhaps it is too simple.

The following is a more difficult version that I did not send- because of the difficulty.

It is of course a better problem.

beg pie
whi roy sf3
bla kh6 qc4 rh2 bd2g4
stip ser-#7
cond whitemaxi chameleonch sentinelles pionadvers supercirce
twi mov c4 a2
end

Kjell Widlert
March 9, 2015 21:46

I hadn’t noticed that the reborn bQ must also guard a1 to make 6.rBxh2 legal. Now I know!

Yes, the version is certainly a much better problem and it’s the one I would have chosen to publish, had I been the composer. I especially like the successive rebirths of the bR in part b): on g3, e6 and finally c4 – instead of sending it directly to c4!

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