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StrateGems

No. 10, Mar-Jun 2000

R0040 - Tivadar Kardos & ?#8364;rpàd Molnàr

StrateGems 10, Mar-Jun 2000

Black has no legal last move, so he has the move. Mate in 1 with

0. ...Bf4/Be7/Nf6/Qf8/Qg7/Qh5/Qg6/Qh7/ab6/ba6/bc6/cb6/Nb6

1. Nf6/Re7/Nf6/ef8=[QR]/Ng7/Ng7/Bg6/gh7/Ra8/Qb8/Qc6/Nd6/Nc7#


R0041 - A.A. Kislyak

StrateGems 10, Mar-Jun 2000

The last moves were: -1. .. b6xNa5 -2. Nc4-a5 e6-e5 -3. Ne5-c4 f7xBe6 -4. Ra5-a6 Kc8-b8 -5. a6xRb7, and the position unlocks. White captured e2xd3xc4xb5xPa6. The captures on d3, c4, and b5 aren't uniquely determined, but since the black a-pawn was captured on his own file, the capture on a6 must be of this pawn. This explains too why white must uncapture a rook on b7: If white uncaptures it on d3, c4 or b5 it can't get back to a8 or h8.

So the last four captures were: b6xNa5, f7xBe6, a6xRb7 and b5xPa6.


R0042 - ?#8364;rpàd Molnàr

StrateGems 10, Mar-Jun 2000

The last moves were: -1. .. h4xNg3 -2. Rf6-g6 g6xNh5 -3. Rf8-f6 f7xNg6 -4. Bf6-g5 g5xNh4 -5. Nh8-g6 g6-g5 -6. Nf3-h4 Kh4-g4 -7. Ng5-f3. The four promoted knights all originated on d8.


R0043 - Nikita Plaksin

StrateGems 10, Mar-Jun 2000

The intention was: -1. .. e2xQd1=B(+bBc8) -2. Qh1xRg2(+wQd1).

But this is cooked as Michel Caillaud showed: -1. .. e2xQd1(+bBc8) -2. Qh6xBf8(+wQd1), or -1. .. Qa1-a4 -2. Kb4-a5 c4xb3(+bPb7!)


P0054 - Andrei Frolkin & Valeri Gorbunov

StrateGems 10, Mar-Jun 2000

a) 1. a4 Nc6 2. a5 Ne5 3. a6 Ng4 4. ab7 Nh2 5. b8=N Nf1 6. Nc6 dc6 7. g3 Bh3 8. Rh3

b) 1. a4 Nc6 2. a5 Na5 3. g3 Nc4 4. Bh3 Ne3 5. Bd7 Nf1 6. Bc6 bc6 7. h3 Bh3 8. Rh3


P0055 - Andrei Frolkin & Mikhail Kozulya

StrateGems 10, Mar-Jun 2000

Dedicated to Thomas Volet

1. e3 h5 2. Qg4 hg4 3. Ne2 Rh3 4. Rg1 Rf3 5. gf3 e5 6. Rg2 e4 7. fe4 d5 8. ed5 Nc6 9. dc6 Bb4 10. cb7 Bc3 11. bc3 Qd4 12. cd4 Bf5 13. Nc3 Kd7 14. Rb1 Re8 15. Rb2 Re5 16. de5 Nf6 17. ef6 Bh7 18. fg7


P0056 - Michel Caillaud

StrateGems 10, Mar-Jun 2000

1. a4 e5 2. a5 e4 3. Ra4 e3 4. Re3 Ne7 5. b4 Nc6 6. Bb2 Nd4 7. Qc1 Ne2 8. Bf6 Ng3 9. d4 Ngf5 10. Ke2 Nh6 11. Kf3 Nhg8 12. Kg4 e2 13. Qh6 e1=N 14. Ne2 Nd3 15. Nc1 Nc5 16. Bd3 Na6 17. Rhe1 Nb8 18. R1e3 a6 19. Re7


P0057 - Mark Kirtley

StrateGems 10, Mar-Jun 2000

1. h4 Nf6 2. Rh3 Ne4 3. Rc3 Nd2 4. Nf3 Ne4 5. Qd6 Nc6 6. Qa3 d6 7. Bf4 d5 8. e3 d4 9. Bb5 d3 10. Be5 d2 11. Ke2 d1=R 12. Nbd2 Rh1 13. Rd1 Rh3 14. Nb1 Rg3 15. Nfd2 Rg6 16. Kf3 Re6 17. Kg4 Rd6 18. Kh5 Rd7 19. g4 Nb8 20. Bd7


P0058 - Thierry le Gleuher

StrateGems 10, Mar-Jun 2000

1. Nc3 h5 2. Ne4 h4 3. Ng3 hg3 4. Nf3 gh2 5. Rg1 hg1=B 6. c3 Bh2 7. Qa4 Bf4 8. Qa7 Be3 9. Qd4 Ra4 10. Qg4 Rc4 11. a4 Ba7 12. a5 b6 13. a6 Bb7 14. Ra5 Be4 15. Rh5 Nc6 16. Rh1 Bb8 17. a7 Bh7 18. a8=R g6 19. Ra3 Bh6 20. Rb3 Be3 21. de3 Kf8 22. Nd2 Kg7 23. Nb1


P0059 - Unto Heinonen

StrateGems 10, Mar-Jun 2000

The intention was: 1. a4 c5 2. a5 c4 3. a6 c3 4. ab7 a5 5. g4 Na6 6. b8=Q cb2 7. Qe5 d6 8. c4 Be6 9. c5 Bb3 10. c6 e6 11. c7 Ne7 12. c8=Q Rg8 13. Qcc3 Nc6 14. g5 Qf6 15. g6 OOO 16. gh7 g5 17. e3 g4 18. Bb5 g3 19. Ne2 g2 20. Rf1 g1=Q 21. Ra4 Qgg7 22. Rg4 a4 23. d4 a3 24. d5 a2 25. Qdd4 a1=Q 26. Nd2 b1=Q 27. h8=Q Qbb2

This problem is cooked: 1. g4 c5 2. g5 c4 3. g6 c3 4. a4 cb2 5. gh7 g5 6. a5 g4 7. a6 g3 8. ab7 a5 9. c4 Na6 10. b8=Q g2 11. Qe5 d6 12. c5 Be6 13. c6 Bb3 14. c7 e6 15. c8=Q Ne7 16. Qcc3 Nc6 17. e3 Qf6 18. Bb5 OOO 19. Ne2 Rg8 20. Rf1 g1=Q 21. Ra4 Qgg7 22. Rg4 a4 23. d4 a3 24. d5 a2 25. Qdd4 a1=Q 26. Nd2 b1=Q 27. h8=Q Qbb2, with many possibilities to interchange two moves (for instance white's fourth and fifth)


H0313 - Dr. Tomislav Petrovic

StrateGems 10, Mar-Jun 2000

If black has the move in the diagram position, what was white's last move? Because white has 8 pawns, Bh6 is the original bishop from c1, so the last move wasn't d2-d4. It can't have been c3xd4 either, because this would imply 4 white captures, one too many. So white's last move was f2-f4.

If white has the move in the diagram position, what was black's last move? His pawn structure requires 4 captures, and none of them could've been on the last move. So the only possible last move is b7-b5.

h#1 with 1. gf3 Qf3 and duplex with 1. ab6 ab6#.