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Probleemblad

September 2003

R210 - Thierry le Gleuher

Probleemblad, Sep. 2003

-1... OOO -2. Kg7-h8 f7-f6 -3. Kf6-g7 e4-e3 -4. Kf5-f6 e5-e4 -5. Ke4-f5 e6-e5 -6. Kd3-e4 e7-e6 -7. Re6-e2 e2-e1=B -8. Qb2-a1 e3-e2 -9. Ke2-d3 e4-e3 -10. Qc1-b2 e5-e4 -11. Qd1-c1 d6xBe5 -12. Bb2-e5 c6-c5 -13. Bc1-b2 c7-c6 -14. b2-b4 b3xRa2 -15. Ra1-a2 b4-b3 -16. Ba2-b1 b5-b4 -17. Bb3-a2 b6-b5 -18. Ba4-b3 b7-b6 -19. Bb5xPa4 a5-a4 -20. Ke1-e2 a6-a5 -21. Bf1-b5 a7-a6 -22. e2xNf3


R211 - Michel Caillaud

Probleemblad, Sep. 2003

Dedicated to Guus Rol

The repetition cycle must be shorter than 50 half-moves, to avoid premature ending due to the 50-move rule. The shortest repetition cycle is: 1... Kc4+ 2. Bc5 Bc8 3. Bd4 Bg4 4. Bc5 Bf3 5. Bd4 Rg4 6. Bc5 Rg3 7. Bd4 Rh3 8. Bc5 Rh1 9. Bd4 Rb1 10. Bc5 Rb4 11. Bd4 Kb3 12. Bc5 Rd4 13. Bf8 Kc4 14. Bg7 Kc5 15. Bh8 Rb4 16. Bg7 Rb1 17. Bh8 Rh1 18. Bg7 Rh3 19. Bh8 Rg3 20. Bg7 Rg4 21. Bh8 Rf4 22. Bg7 Bg4 23. Bh8 Bc8 24. Bg7 Ba6 25. Bf8+, which is exactly 48 half-moves. This cycle occurred two times. So the position 96 half-moves ago is the diagram position. What was white's last move before this? It must've been Bg7-f8. But wouldn't that imply a premature repetition of the diagram position with Bf8 on g7? Not if the black move before that was e7-e5, with an en-passant option!


R212 - Michel Caillaud

Probleemblad, Sep. 2003

Last move was Rh7-h6. This position can't be repeated, so the 50-move rule must be applied. Because white had the last move in this sequence, the last pawn move must've been white's too. The only possibility is e6-e7. So the last 101 halfmoves were (start from the diagram, with Pe7 on e6, Kg2 on d3 and Rh6 on h7):

1. e7 Qh6 2. Kc4 Qh5 3. Kd3 Qh6 4. Kc4 Qh5 5. Kb4 Qh6 6. Ka3 Qh5 7. Kb4 Qh6 8. Ka3 Qh5 9. Ka2 Qh6 10. Kb1 Qh5 11. Ka2 Qh6 12. Kb1 Qh5 13. Kc1 Qh6 14. Kd1 Qh5 15. Kc1 Qh6 16. Kd1 Qh5 17. Ke1 Qh6 18. Kf1 Qh5 19. Ke1 Qh6 20. Kf1 Qh5 21. Kg2 Qh6 22. Kh1 Qh5 23. Kg2 Qh6 24. Kf3 Qh5+ 25. Ke3 Qh6+ 26. Kf3 Qh5+ 27. Ke3 Qh6+ 28. Kd3 Qh5 29. Kc4 Qh6 30. Kd3 Qh5 31. Kc4 Qh6 32. Kb4 Qh5 33. Ka3 Qh6 34. Kb4 Qh5 35. Ka3 Qh6 36. Ka2 Qh5 37. Kb1 Qh6 38. Ka2 Qh5 39. Kb1 Qh6 40. Kc1 Qh5 41. Kd1 Qh6 42. Kc1 Qh5 43. Kd1 Qh6 44. Ke1 Qh5 45. Kf1 Qh6 46. Ke1 Qh5 47. Kf1 Qh6 48. Kg2 Qh5 49. Kh1 Qh6 50. Kg2 Qh5 51. Rh6


R213 - Paul Raican

Probleemblad, Sep. 2003

Intention: 1. d4 e5 2. Bf4 exd4 3. Bxc7 Qxc7 4. Qxd4 Qxc2 5. Qxg7 Qxb2 6. Qxg8 Qxa2 7. Qxh7 Qxb1+ 8. Qxb1 Rxh2 9. Qxb7 Rxg2 10. Rh7 Rxg1 11. Rxf7 Kxf7 12. Qxc8 Ke6 13. Qxf8 Kd5 14. Qxb8 Kc4 15. Qxa7 Kb3 16. Qxd7 Rxa1+ 17. Qd1+ Ka2

But Dolf Wissmann found a cook: 1. d3 h6 2. Bxh6 Nc6 3. Bxg7 Rxh2 4. Bxf8 Rxg2 5. Bxe7 Kxe7 6. Rh7 Kd6 7. Rxf7 Kc5 8. Rxd7 Rxg1 9. Rxc7 Qxd3 10. Rxb7 Qxc2 11. Rxa7 Kb4 12. Rg7 Rxa2 13. Rxg8 Rxb2 14. Rxc8 Rxb1 15. Rxc6 Rxa1 16. Rxc2 Kb3 17. Ra2+ Kxa2


R214 - Unto Heinonen

Probleemblad, Sep. 2003

1. d3 h6 2. Bxh6 g6 3. Bc1 Rxh2 4. g4 Rg2 5. Rh8 Bh6 6. g5 Kf8 7. Kd2 Kg7 8. Kc3 Nf6 9. Kb4 Qg8 10. Ka5 Qh7 11. Rxc8 Na6 12. Rh8 Qg8 13. a4 Qd8 14. Ra3 Ng8 15. Rc3 Kf8 16. Rc4 Ke8 17. c3 Bf8 18. Rh1 Rh2 19. Bg2 Rh8 20. Nh3 Rc8 21. Qg1 Nb8 22. Bf1 a6

Nine switchbacks without promotions


R215 - Hans Uitenbroek, Koen Versmissen, Joost de Heer

Probleemblad, Sep. 2003

1. g3 a5 2. Bh3 Ra6 3. Kf1 Rh6 4. Kg2 g6 5. Kf3 Bg7 6. Ke4 Kf8 7. Kd5 Qe8 8. Kc5 Rh5+ 9. Kb6 Qd8+ 10. Ka7 Ke8 11. Bf1 Bf8

Four switchbacks in a capture-free game.