No.1185 
Original Fairy problems 
Definitions: (click to show/hide)
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.
Circe Equipollents: After a piece is captured, it is immediately replaced on the square which is the same distance and direction from the square of its capture, as was that square from the square upon which its captor commenced its move. (If Qf7 captures a Pawn on e7, it is reborn on d7, because d7 is a same distance and direction from e7 as e7 is from f7. Similarly, if Qg7 captures a piece on ‘e7’ its rebirth square is ‘c7’). If the rebirth square is occupied the captured piece disappears
Minimummer: Black must play his geometrically shortest move or may choose from among shortest 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 Pythagoras’s s theorem (take the square root of the sum of the squares of the orthogonal distances). All other orthodox chess rules apply. White Minimummer: White must make moves with minimal possible length.
Marguerite (MG): Moves like all simple hoppers (eagle + grasshopper + hamster + moose + sparrow).
Grasshopper (G): Moves along Qlines over another unit of either color to the square immediately beyond that unit. A capture may be made on arrival, but the hurdle is not affected.
Eagle (EA): A grasshopper which pivots 90^{ o} (to either side) at the hurdle.
Hamster (HA): Moves like a Grasshopper but deflect 180^{o} on passing over the hurdle.
Moose (M): A grasshopper which pivots 45^{ o} (to either side) at the hurdle.
Sparrow (SW): A grasshopper which pivots 135^{ o} (to either side) at the hurdle.
No.1185 Erich Bartel 
Solution: (click to show/hide) 
white kh1
black gh3 mgh4
hs#10* (1+2) 
1...MGh4g2 2.Kh1g1 Gh3f1 + 3.Kg1*f1[+bGe1] MGg2e2 + 4.Kf1g1 Ge1h1 5.Kg1g2 MGe2h3 + {
} 6.Kg2f2 Gh1h4 7.Kf2f1 MGh3e1 8.Kf1f2 MGe1g3 + 9.Kf2f1 Gh4f2 #
1.Kh1g1 MGh4g2 2.Kg1h1 Gh3f1 3.Kh1h2 Gf1h3 + 4.Kh2*h3[+bGh4] MGg2g4 + 5.Kh3h2 Gh4h1 {
} 6.Kh2g2 MGg4f1 + 7.Kg2g3 Gh1e1 8.Kg3h3 MGf1h4 9.Kh3g3 MGh4f2 + 10.Kg3h3 Ge1g3 # {
(C+ by Popeye 4.75)}

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