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1 Rf7 Bb4 (1.... Bg7 2. Rb8) 2 B7 g6
Does my idea (post 48) works or fails
It fails, there is ONLY ONE answer, confirmed by an engine.
Rf7 fails, I confirm, no need of engine.
Ps: it can be solve by commun engine ;)
but if 1..... be7...then stalemate may arise on 2 r x e7
1.bg8 bg7 (to avoid Rh7#) 2.Rb1+ bf8 3.Rh7# am i right?
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