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d3;
august 25, 2009 - 8:12 pm1.Ta2!
1…Na6 2.Ta6 Na7(2…ba6 3.Df3 Ra7 4.Db7#) 3.Ta7! Ra7 4.Da4#;
1…Na7 2.Ta7! Ra7 3.Da4 Na6 4.Dd4 Ra8(4…b6 5.Db6 Ra8 6.Da6#)5.Dh8 Ra7 6.Db8#
d2:
august 25, 2009 - 8:42 pm1.Td8! Rb7 2.Dc6!! Rc6 3.Ne4 Rc5 4.Td5 Rc6 5.Td4 Rc5 6.b4#
Virgil, corect solutiile. Felicitari!
august 25, 2009 - 9:41 pmD1: 1.C:b6+ a:b6 [1…T:b6 2.Dh1+ Tb7 3.Dh8+ Tb8 (3…Df8 4.Td8+ D:d8 (4…Tb8 5.Dh1#) 5.D:d8 Tb8 6.D:b8+# ) 4.D:b8+ R:b8 5.Td8+ Rb7 6.Tb8# ] 2.Da2+ Ta7 3.Dg8+ Rb7 4.Db8+ R:b8 5.Td8+ Rb7 6.Td8
august 26, 2009 - 11:23 amD4: 1.e4 Rh6 2.Cef8 Rh5 3.Rg7 Rg5 4.Ch7+ Rh5 5.Cf6+ Rg5 6.h4+# Probabil ca albul poate inversa primele 2 mutari 1.Cef8 Rh6 2. e4 Rh5 si se ajunge la acelasi rezultat
Manolache Marius:
august 26, 2009 - 9:52 pmcorect solutiile la diagrama 1 (o mica greseala de notare 6.Te8) si 4. Felicitari!