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
October 23, 2019 23:26

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
October 26, 2019 23:32
Reply to  Kjell Widlert

Could this be fixed by mirroring the position vertically?

Joost
Joost
October 27, 2019 12:14
Reply to  Joost

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

Paul Rãican
Paul Rãican
October 28, 2019 11:55

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

Joost
Joost
October 28, 2019 12:44
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
October 28, 2019 13:48

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

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