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Awards Schachmaty w SSSR

1985 - Solutions

1 - A. Kisljak

Black is mated, but is the position legal? Yes, and the 35 last single moves are determined.

Retract -1. Nb5-a3# Re8-d8 -2. e5-e6 Rd8-e8 -3. e4-e5 Re8-d8 -4. e3-e4 Rd8-e8 -5. e2-e3 Re8-d8 -6. Qe3-e7 Re7-e8+ -7. Qh6xPe3 e4-e3 -8. Qh8-h6 e5-e4 -9. h7-h8=Q e6-e5 -10. g6xRh7 Rh6-h7 -11. g5-g6 Rg6-h6 -12. g4-g5 Rg5-g6 -13. g3-g4 Rc5-g5 -14. Bb6-c7 Rc7-c5+ -15. Be3-b6 h6-h5 -16. Bc1-e3 h7-h6 -17. d2xNc3 Kd4-c4 -18. Na3-b5+.


2 - F. Fatchullin

Black has the move, so that 1 ... Dxf1# and not 1. Ne4?.

Retract -1. Qg1-f1 Qf1-e1 -2. Kg2-h1 Qe1-f1+ -3. Nh1-f2 b3-b2 -4. Qb6-g1 b4-b3 -5. Qb8-b6 b5-b4 -6. b7-b8=Q b6-b5 -7. a6xRb7 Re7-b7 -8. Nf2-h1 Re4-e7 -9. Nh1-f2 Rd4-e4 -10. Nf2-h1 Rf4-d4 -11. Nh1-f2 Rf1-f4 -12. a5-a6 f2-f1=R -13. a4-a5 e3xBf2 -14. Bg1-f2 e4-e3 -15. Bc5-g1 e5-e4 -16. Ba3-c5 e6-e5 -17. Bc1-a3 b6-b7 -18. b2xNc3 and the position unlocks


3 - A. Kornilov

The position is legal and can be unlocked by uncapturing 5 bl. units on c7-...-g7. Only possibility is to uncapture h6xPh7, g6xNf7, f6xBe7, e6xRd7 & d6xQc7 (in that precise order).


4 - B. Lurje and N. Plaksin

The 6x8 part (call it A) can be reversed (call this a) and similarly with the 2x8 part, yielding B and b. Thus there are eight different boards that can be arranged: AB, Ab, aB, ab, BA, bA, Ba, ba.

Only the aB arrangement gives a legal position with a mate in one presentation. There Black has the move so that he plays 0 ... Rxf5/Dxf5 to which White replies 1. Ne7/Bxf5 #