No.1439
Andreas Thoma (Germany)

Original Retro & PG problems
JF – R2019-20


Definitions: (click to show/hide)


No.1439 Andreas Thoma
Germany

original – 13.09.2019 
Klaus Wenda zum 78. Geburtstag gewidmet

Solution: (click to show/hide)

white Ke3 Qc4 Pa6c2e2g2 Rh8 black Ke5 Qb5 Sb1h1 Bd6e4 Rc3h3 Pc5e7g4h5

-5 & s=1                                   (6+12)
Proca Retractor
Anti-Circe Cheylan


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Kjell Widlert
Kjell Widlert
11 months ago

Isn’t this cooked (dualized) by (1.h4xBg5>g2 Bf6-g5+) 2.Kd3-e3 Ba8-e4+ (to cancel checks from Be4, Rc3, and Rh3) 3.Ke3-d3 & forwards 1.Qe6+ Kxe6# (legal as Rxe8?? is illegal due to Sh1; stalemate as Rh8 is pinned by Rc3 and Rh3) ?
Or am I overlooking something?

Joost
Joost
10 months ago
Reply to  Kjell Widlert

Could this be fixed by mirroring the position vertically?

Joost
Joost
10 months ago
Reply to  Joost

No, because 1. Qd4 Bxd4[Bc8] is also possible as s=1.
Perhaps +wPa6.

Paul Rãican
Paul Rãican
10 months ago

It isn’t possible 1.Qd4+ as s=1, because Black has 1. …Kc6!

Joost
Joost
10 months ago
Reply to  Paul Rãican

Perhaps I wasn’t clear: Mirroring the position doesn’t work, because Kjell’s dual would still work: 1. a4xBb5->b2 Bc6-b5+ 2. Ke3-d3 Bh8-d4+ 3. Kd3-e3 and 1. Qd4 Bxd4->c8=

Paul Rãican
Paul Rãican
10 months ago

Right! On my board, a black Bishop c6 was missing.