# No.1293 (HG)

 No.1293 Hubert Gockel (Germany) Original Fairy problems JF-10/2017-3/2018: October’2017 – March’2018

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 No.1293 Hubert Gockel Germany original – 30.03.2018 Solution: (click to show/hide) white Qa3 Ba8 Re5f8 Sg6 Bg3 Kg8 Pb6c7f5 black Kd6 Sb5c4 Qc2 Pc5d7e6 #2                                           (10+7) AMU 1.R:e6#?? / Q:c5#?? / c8=S#?? These thematic mates are illegal because each mating piece is observed twice. In each of three defences Black withdraws two observations of two pieces in cyclic fashion so that two mates mates should be possible at the same time. But thanks to cyclic dual avoidance only one is. (Author) 1.Sg6-e7 ! threat: 2.Se7-c8 # 1...Sc4*b6 2.Re5*e6 # {A (2.Q:c5+? B Q:c5!)} 1...Sb5-a7 2.Qa3*c5 # {B (2.c8=S+? C S:c8!)} 1...Kd6*e7 2.c7-c8=S # {C (2.R:e6+? C d:e6!) (C+ by Popeye 4.79)}

### 6 Responses to No.1293 (HG)

1. Instructive example of cyclic dual avoidance and a new proof of AMU condition usability for various effects.

2. Geoff Foster says:

An interesting problem with a good flight-giving key. The dual avoidance uses 3 different mechanisms: opening of black line, guard of mating square, and move of King so that white move is not a double-check.

3. Geoff Foster says:

However it is a pity that after 1…Kxe7, 2.Qxc5 would not be mate anyway.

• seetharaman says:

Can d7 be shifted to f7?

4. Geoff Foster says:

@Seetharaman: I don’t understand what you mean about shifting d7 to f7. The bPd7 is necessary in order to block the square d7.

• seetharaman says:

I meant if a suitable guard could be found for d7.