ALL DE FR ES IT

Logo

Europe Echecs

No. 449 , Oct. 1996 - SOLUTIONS

626 - A. Zolotarev
Europe Echecs 449, 10/1996

(a) Wh. caps are d2xNe3 & g2xBf3. Bl. caps are c7xb6, d7xc6, exd, hxg, gxf & wPa2 on its file. wPh2 promoted into a wBh8. Retract:
-1. c7xNb6+ b4-b5
-2. g5xNf4 Nd3-f4
-3. Ra2-b2 Nb2-d3+
-4. Ra4-a2 b3-b4
-5. Rb4xPa4! Nc4-b6+
-6. Rb6-b4+ a3-a4
-7. d3-d2 Nd2-c4+
-8. e4xBd3 Bc4-d3
-9. e5-e4 Be6-c4
-10. f7-f6 Bh3-e6
-11. e6-e5 Bf1-h3
-12. e7-e6 g2xBf3
-13. Bg4-f3 a2-a3
-14. Bc8-g4 Kd5-c5
-15. d7xRc6+ and the position unlocks.

(b) Wh. caps are d2xNe3, g2xBf3 and bPa7 on its file. Bl. caps are c7xb6, d7xc6, exd, hxg & gxf. wPh2 promoted into a wBh8. Retract
-1. c7xNb6+ a2-a3
-2. g5xNf4 Nd3-f4
-3. Rb3-b2 Nb2-d3+
-4. Ra3-b3 b4-b5
-5. Ra6-a3 Nc4-b6
-6. Rb6-a6+ Na3-c4
-7. f7-f6 Nc4xPa3!
-8. d3-d2 Nd2-c4+
-9. e4xBd3 Bc4-d3
-10. a4-a3 Be6-c4
-11. e5-e4 Bh3-e6
-12. e6-e5 Bf1-h3
-13. e7-e6 g2xBf3
-14. Bg4-f3 b3-b4
-15. Bc8-g4 Kd5-c5
-16. d7xRc6+ etc.


627 - T. Le Gleuher
Europe Echecs 449, 10/1996

The position can be reached e.g. with the following proof game:

1. d3 Nf6 2. Bf4 Nd5 3. Qd2 Nb6 4. Qe3 Nc6 5. Nd2 Rb8 6. O-O-O e5 7. Qh3 e4 8. Kb1 e3 9. Ka1 exd2 10. e4 f5 11. Be5 f4 12. Qe3 fxe3 13. Bg3 g5 14. f4 g4 15. Re1 Na8 16. Re2 b6 17. Rf2 e2 18. Rf3 Ne5 19. Re3 Kf7 20. Nf3 gxf3 21. Be1 Be7 22. Kb1 Bh4 23. Ka1 Bf2 24. Rg1 Kg6 25. Kb1 Kh5 26. Ka1 Kh4 27. Kb1 Bb7 28. Ka1 Bd5 29. Kb1 Be6 30. Ka1 Bh3 31. g4 h5 32. Bg2 Qf6 33. Bh1 Bf1 34. h3 Rh7 35. Bg2 Kg3 36. Bh1 Kh2 37. Kb1 h4 38. Rg3 hxg3 39. h4 Rg7 40. h5 Kg1 41. h6 Rbg8 42. h7 Kh2 43. h8=R Kg1 44. Rh7 Bg2 45. Rh2 Kf1 46. Ka1 Bh3 47. Rg2 Bg1 48. Kb1 Bh2 49. Rf2 Kg1 50. Ka1 Bf1 51. Bg2 Qh6 52. Bh3 Kh1 53. Kb1 Bg1 54. Ka1 Bg2 55. Rf1 Kh2 56. Bf2 Kh1 57. Rc1 Kh2 58. Rb1 Bh1 59. Bf1 Qh3 60. Be1 Qg2 61. Rc1 Qf2 62. Bh3 Qf1 63. Bf2 Qd1 64. Bf1 Rh7 65. Be1 Bg2 66. Rb1 Kh1 67. Rc1 Rh2 68. Rb1 Bh3 69. Rc1 Rf2 70. Rb1 Kh2 71. Rc1 Bg2 72. Rb1 Bh1 73. Bh3 Rf1 74. Bf2 Re1 75. Bf1 Bg2 76. Rc1 Kh1 77. Rb1 Rh8 78. Rc1 Rh2 79. Rb1 Bh3 80. Rc1 Rg2 81. Rb1 Bh2 82. Rc1 Bg1 83. Rb1 Bh2 84. Bg1 Qc1 85. Bf2 Rd1 86. Be1 Ng6 87. Bf2 Bg1 88. Be1 Rf2 89. Bg2 Kh2 90. Bh1 Rf1 91. Bf2 Bg2 92. Be1 Kh3 93. Bf2 Kh4 94. Be1 Bh3 95. Bg2 Bh2 96. Bf2 Rg1 97. Bh1 Bf1 98. Be1 Kh3 99. Bf2 Nh4 100. Be1 Ng2 101. Bf2 Ne1 102. Bg2 Kh4 103. Bh1 Bh3 104. Bg2 Rf1 105. Bg1 Rf2 106. Bf1

where the last pawn-or-capture move is 43. h8=R, more than 50 moves ago. I did not try to find a shorter way of reaching the position....


628 - A. Zolotarev
Europe Echecs 449, 10/1996
(This problem was later found to have duals)

Bl. seven caps. are cxdxexfxg, dxe, g7xf6 and -1. Bh3xg4+. This requires 5 Wh. promotions: Ph2 on g8, Pg2 on g8, Pa2 on b8, Pb2 on c8, Pc2 on c8 (accounting for all 3 caps). The promoted wB is from b8.

It is possible to have a shortest number of Q moves by reconducting the bQe3 on d8 in just two moves, and by assuming no Q-promotion. E.g. by retracting
-1. Bh3xNg4+ Nh6-g4
-2. g4-g3+ Ng8-h6
-3. h7-h5 g7-g8=N
-4. d6xRe5 h6xBg7
-5. Bf8-g7 Td5-e5
-6. Rg8-g6 Be5-h2
-7. Qb6-e3 Bd4-e5
-8. Qd8-b6 Bb6-d4
-9. Rh8-g8 Bc7-b6
-10. ... Bb8-c7
-11. ... b7-b8=B
-12. ... a6xNb7
-13. ... Rh5-h4
-14. ... Rg5-h5
-15. ... Rg8-g5
-16. ... g7-g8=R
-17. ... g6-g7
-18. ... Kg5-f5
-19. g7xR/Nf6+ and everything unlocks.

Author's intention was perhaps something like
-1. Bh3xNg4+ Nh6-g4
-2. Rg4-g6+ Ng8-h6
-3. d6xRe5 Rc5-e5
-4. f4xNg3 Rc8-c5
-5. e5xNf4 c7-c8=R
-6. d7-d6 c6-c7
-7. Qb6-e3 c5-c6
-8. d6xRe5 b4xBc5
with more determined promotions (wNg8 and wRe5 on g8) and uncaptures but there is in fact no way of further unlocking because the bK/bQ must reach e8/d8 before we free the wK with g7xf6 and c7xd6.


629 - A. Kornilov
Europe Echecs 449, 10/1996

Add a Rh8 and paint Bl. the following Ba8, Pa7, Pb6, Pc6, Pe6, Pf6, Rf3, Pg7, Ph7, Rh8, Kh6, yielding

[bB4Br/pPP1PPpp/1pp1ppPk/2P2P1Q/3R2K1/4Nr1P/5N2/6R1]

16+11.

[bB4Br/pPP1PPpp/1pp1ppPk/2P2P1Q/3R2K1/4Nr1P/5N2/6R1]

Here, Wh. caps are a6xb7, b6xc7, d6xe7, e6xf7 & last move Qxh5 mate. For balance, the bPd7 promoted on d1.

Last three moves are -1. Qh4xNh5 Nf4-h5 -2. Qg3-h4+. It is necessary to add a bR precisely on h8 because unlocking requires reconducting bBf8 and bPe7 before we can extract the wBg8. Also, it is not possible to exchange Ba8 and Bg8 because then it becomes impossible to reconduct the bQR on a8.