Rook-Hopper(RH): (0,1) Hopper. Moves along Rook lines 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.
Bishop-Lion(BL): (1,1) Lion. Moves along Bishop lines over another unit of either color to any square beyond that unit. A capture may be made on arrival, but the hurdle is not affected.
No.1289.1Francesco Simoni Italy version of No.1289 – 13.06.2020
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
a)
1...Bc8-b7 2.Kd5-d4 Sb2-a4 3.Sc6-e5 Sg7-f5 #
b) bRHd8-->a5
1...Bf1-g2 2.Kd5-e5 Sg7-e8 3.Sf3-d4 Sb2-d3 #
c) bRHe1-->a5
1...Bc8-b7 2.Kd5-d4 Sb2-d1 3.Sc6-e5 Sg7-e6 #
d) bRHh4-->a5
1...Bf1-g2 2.Kd5-e5 Sg7-h5 3.Sf3-d4 Sb2-c4 # {
(C+ by Popeye 4.83)}
In a) it is possible to move 2.Sb2-a4. In each twin, instead, one of the Rook-Hoppers is
moved to a5 in turn, preventing so the move 2.Sa4 because it would be illegal.
However, in b) the mate in d3 becomes possible because the RH moved in a4 no longer
guards d3. in c) the move Sb2-d1 becomes legal and in d) the mate in e4 becomes
possible because the RH moved in a4 no longer guards e4. (Author)
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!)
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).
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!)
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).
Have corrected the stipulation under diagram to h#2.5. Thank you, Kjell!
Compared to 1289,
Pros: piece economy(aristocrat Meredith!), 4 solutions
Cons: twinning(instead of multi-solutions), more obvious symmetry, 6 repeated moves
Everything is shareable. In practice, the reasons why I choose the version in 3 moves.