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Original Fairy problems |
No.1197
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No.1197 Aleksey Oganesjan |
Solutions: (click to show/hide) |
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white Kg1 Bb2
black Kd3 Rd2 Pb3c2
h=4 (2+4) |
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Original Fairy problems |
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No.1197 Aleksey Oganesjan |
Solutions: (click to show/hide) |
|
white Kg1 Bb2
black Kd3 Rd2 Pb3c2
h=4 (2+4) |
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Very original, I think! Shnoebelen theme brought into normal play (ie other than Retro/PG!
Nitpick: It’s ‘Schnoebelen’.
An economical way of forcing three different wK routes, in turn forcing three different Schnoebelen promotions on c1. Of course, it is impossible to force a Schnoebelen queen without some additional condition…
This is basically a ser-#4:
W: Kg1
B: Ka1 Rd2 Pa2 Pc1
ser-#4
(b) Rd2->e2
(c) Rd2=Bd2
(a) Pawn c1 promotes to a knight, then 1. Kf1 2. Ke1 3. Kd2 4. Kc1=
(b) Pawn c1 promotes to a bishop, then 1. Kf1 2. Ke2 3. Kd1 4. Kc1=
(c) Pawn c1 promotes to a rook, then 1. Kf2 2. Ke2 3. Kd2 4. Kc1=
Probably you mean ser-=4 (not ser-#4)?
It looks like a joke problem which usually has a stipulation with 1/2-move, 1/3-move, 1/4-move and etc :)) Such problems often includes unusual chess moves – castling, promotion and en-passant-capture – that consist of some “components”.
In your version the Pawn already executed a move c2-c1 (or b2xc1) but does not yet promote in any piece – so, it would be really say that the Pawn executed 1/2-move 🙂
I think my h=4-problem is “not fully serial”: yes, Black can execute their moves serially regardless of White (c1X-Kc2-Kb1-Ka2) but White cannot execute their moves serially (Kg1-f~-e~-d~xc1) because at first bK must take off control on some squares that need for wK. So I think that the stipation h= is more suitable than ser-= with “jesting” condition like “Completion of bP promotion” 🙂
Yes, I meant ser-=4.