No.1289.1
Francesco Simoni
(Italy)

Original Fairy problems
JF-2020/I:
January – June’2020


Definitions: (click to show/hide)


No.1289.1 Francesco Simoni
Italy

version of No.1289 – 13.06.2020

Solutions: (click to show/hide)

black RHd8h4e1 BLh8 Sf3c6 Kd5 white Bf1c8 Ka1 Sb2g7

h#2.5                                        (5+7)
b) RHd8→a5 ; c) RHe1→a5 ; d) RHh4→a5
Bishop-Lion h8
Rook-Hopper d8,h4,e1


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Kjell Widlert
Kjell Widlert
3 months ago

The stipulation is h#2.5 (not h#3).
The author’s comment is slightly inaccurate: the reason solution a) does not work in b-d) is not that 2.Sa4 is illegal (a5 is a rook-hopper, not a rook-lion); the real reason is that RHa5 guards the mating square f5. So the logic is the same in all four twins.
(No serious complaint: everybody must have seen this!)

Francesco Simoni
Francesco Simoni
3 months ago
Reply to  Kjell Widlert

You are right sorry. I probably confused with the version published in 2017…
Yes, it’s in 2,5 moves. This is the reason I published a different version, in 3 moves, despite in it one solution was lost (no way I found to avoid a reversal of moves). But I liked to see this version published, even if only in 2.5 moves, because it has four solutions (in twin form).

shankar ram
shankar ram
3 months ago

Compared to 1289,
Pros: piece economy(aristocrat Meredith!), 4 solutions
Cons: twinning(instead of multi-solutions), more obvious symmetry, 6 repeated moves

Francesco Simoni
Francesco Simoni
3 months ago

Everything is shareable. In practice, the reasons why I choose the version in 3 moves.