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| Aleksandr Galitzky, Shakhmatnyy Zhurnal, 1900. Código FEN: 7k/4K1pp/7N/8/8/8/8/B7 w - - 0 1. |
Um problema belíssimo: as Brancas jogam e dão xeque-mate em 3 lances!
(A very beautiful problem: White to play and give checkmate in 3 moves!)
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| Aleksandr Galitzky, Shakhmatnyy Zhurnal, 1900. Código FEN: 7B/8/8/8/8/N7/pp1K4/k7 w - - 0 1. |
Como se não bastasse, giramos o tabuleiro, e novamente: as Brancas jogam e dão xeque-mate em 3 lances!
(As if it were not enough, we rotate the board, and again: White to play and give checkmate in 3 moves!)
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| Aleksandr Galitzky (05/03/1863 - 18/11/1921) |
Este é o autor desta obra de arte: Aleksandr Galitzky, problemista russo, considerado um gênio da composição de xadrez. Hoje ele celebraria seu 154º aniversário.)
(This is the author of this artwork: Aleksandr Galitzky, Russian problemist, regarded as a genius of chess composition. Today he would celebrate his 154th birthday.)
PCFilho
The theme is stalemate avoidance for both parts, as in each case Black has no legal move in the diagrammed position.
ResponderExcluirThe first part is solved by 1 Bf6! (waiting) gxf6 2 Kf8 f5 3 Sf7.
The second part is solved by 1 Kc3! (waiting), with the variations 1 ... b1(Q) 2 Sc2+ Qxc2+ 3 Kxc2 and 1 ... b1(S) 2 Kc2+ Sc3 3 Bxc3. If you count the short mates 1 ... b1(R) 2 Sc2 mate and 1 ... b1(B) 2 Kc2 mate, this part's an Allumwandlung.
As usual, well done, my friend Jake.
ExcluirThank you for posting such a detailed explanation of this beautiful problem/solution. :)
I learn a lot from your solutions.
The solutions:
ResponderExcluirFirst diagram:
1. Bf6!! gxf6 2. Kf8!! f5 3. Nf7#
Second diagram:
1. Kc3!!
1. ... b1=Q 2. Nc2+ Qxc2+ 3. Kxc2#
1. ... b1=R 2. Nc2#
1. ... b1=N+ 2. Kc2+ Nc3 3. Bxc3#
1. ... b1=B 2. Kb3#
Two notes about this amazing problem:
ResponderExcluir- the author's name is sometimes spelled Aleksandr Vasilevich Galickij.
- probably without knowing Galitzky's problem, Georg Albert Becker published, in Badische Neueste Nachrichten, in 1954, a mirrored version of it (FEN: k7/pp1K4/N7/8/8/8/8/7B w - - 0 1).
The solution would be 1. Bc6!! bxc6 2. Kc8!! c5 3. Nc7#
In the second diagram (FEN: B7/8/8/8/8/7N/4K1pp/7k w - - 0 1), the solution would be:
1. Kf3!! g1=N+ (1. ... g1=Q 2. Nf2+ Qxf2+ 3. Kxf2#) 2. Kf2+ Nf3 3. Bxf3#