Peter S. Krug (Аustria)
|
|
1.Qe7! 1.Bb2+? Kh7! 2.Qe7+ Kxh6 3.Bc1+ Kg6= 1.Qe8+? Qg8= 1...c3 1...Qg6+ 2.Ka5! Qg2 (2...c3 3.Bd6 Qxh6 4.Be5+ Kg8 5.Qe8+ Qf8 (5...Kh7 6.Qf7+-) 6.Qg6++-) 3.Qf6+ Kh7 4.Qf7+ Kxh6 (4...Kh8 5.Bf8 Qg5+ 6.Ka4+-) 5.Bc1+- 2.Bd6 Qd4+ 2...Qg6 3.Ka5+- 3.Bc5! 3.Ka5? Qd5+ 4.Bc5 Qa2+= 3...Qg4 4.Ka5! Bc4 5.Bd6 Qg6 6.Be5+ Kg8 7.h7+ Qxh7 8.Qe8#
|
|
|
|
|
