# No.1453 (HG)

 No.1453  Hubert Gockel (Germany) Original Fairy problems JF-2019/II: July – December’2019

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 No.1453 Hubert Gockel Germany original – 20.10.2019 Solution: (click to show/hide) white Qc2 Pc3b6 Sd4 Rf5 Ka1 black Bh1 Ke3 Qa8 Pf3f4e5f7b7a6 Se4 Rf8 #2vv                                          (6+11) Breton Adverse 1.Rf5*e5[-b7] ? threat: 2.Sd4-f5 # {A} 1...Qa8-c8 2.Re5*e4 # {B} 1...f3-f2 2.Qc2-e2 # {C} but 1...f7-f5 ! 1.Rf5*e5[-f7] ? threat: 2.Re5*e4 # {B} 1...f3-f2 2.Qc2-e2 # {C} 1...Rf8-e8 2.Sd4-f5 # {A} but 1...Qa8-e8 ! 1.Rf5*e5[-f3] ! threat: 2.Qc2-e2 # {C} 1...Bh1-f3 2.Sd4-f5 # {A} 1...f4-f3 2.Re5*e4 # {B (C+ by Popeye 4.83)} Cyclic Pseudo-le Grand. (Author)

### 2 Responses to No.1453 (HG)

1. Kjell Widlert says:

The logic is not as simple as one would think (which makes the problem all the more interesting). The Breton adverse condition allows White to remove any black P when capturing Pe5. So White sets up a threat by removing a guarding bP … no, that is true only for the solution! Rf5-e5 actually sets up two threats (Sf5#, Rxe4#), and the bP removals on b7 and f7 serve as dual-avoidance effects (Sushkov-theme style). The removal of Pf3 in the solution rules out both threats by dual-avoidance effects, but lets in a new threat by removal of a guard (so this is a form a threat correction, too).

The appearance of all three thematic mates in every phase works quite seamlessly. I’m especially impressed with how Sf5# and Rxe4# are produced in the solution.

2. Hubert Gockel says:

I couldn’t have explained the contents better – so thank you very much, Kjell!