# No.1478 (GF)

 No.1478  Geoff Foster (Australia) Original Fairy problems JF-2020/I: January – June’2020

Definition: (click to show/hide)

 No.1478 Geoff Foster Australia original – 05.02.2020 Solution: (click to show/hide) white Ka5 Bd1 Se5f5 black Kd5 Qc4 h#8             2 solutions             (4+2) Black Must Check 1.Qc4-c3 + Ka5-a6 2.Qc3-d3 + Se5-c4 3.Qd3-a3 + Ka6-b6 4.Qa3-b2 + Bd1-b3 { } 5.Qb2-f2 + Sf5-d4 6.Qf2-f6 + Sd4-e6 7.Qf6-d4 + Kb6-c7 8.Qd4-e5 + Sc4-d6 # 1.Qc4-c7 + Ka5-b4 2.Qc7-e7 + Sf5-d6 3.Qe7-h4 + Sd6-e4 4.Qh4-e7 + Kb4-c3 { } 5.Qe7-c7 + Se5-c6 6.Qc7-g3 + Bd1-f3 7.Qg3-e5 + Kc3-b4 8.Qe5-d6 + Se4-c5 # { (C+ by Popeye 4.83 & WinChloe 3.47)} A rotated echo with battery cross-check mate. The knights exchange roles. The second solution has switchbacks Qc7-e7-h4-e7-c7 and Kb4-c3-b4. (Author)

### 2 Responses to No.1478 (GF)

1. François Labelle says:

“Black must absolutely check” would be a better translation of BlackUltraSchachZwang. “Black must check” already exists with a different meaning, according to the Problemist Glossary.

Also, the definition of BlackUltraSchachZwang given here is missing the statement “Stalemate and mate remain orthodox” found in the PDB definition. This is important otherwise there are short solutions like 1.Qc7+ Kb4 2.Qe7+ Sxe7#.

2. seetharaman kalyan says:

Seems without the fairy condition also the solution will as it is helpmate.