No.1730
James Malcom
(USA)
Original Fairy problems
JF-2023:
01.01.2023 - 31.12.2023
No. 1730 James Malcom
USA
original - 22.01.2023
White Kc1 Qd7 Sa1b3 Bb8 Pa4b4c2f2h3
Black Kb6 Qc8 Ra5 Bb1 Pb7c3c7f3f6f7h4h7
ep 104 10+12
White forces an en passant capture to take place; C-
White forces an en passant capture to take place; C-
Solution: (click to show/hide)
1. b4*a5+ Kb6-a6 2. Qd7*c8 Bb1-a2 3. Kc1-d1 Ba2-b1 4. Kd1-e1 Bb1-a2 5. Ke1-f1 Ba2-b1 6. Kf1-g1 Bb1-a2 {
} 7. Kg1-h2 Ba2-b1 8. Kh2-h1 Bb1-a2 9. Kh1-g1 Ba2-b1 10. Kg1-f1 Bb1-a2 11. Kf1-e1 Ba2-b1 12. Ke1-d1 Bb1-a2 {
} 13. Kd1-c1 h7-h6 14. Kc1-d1 Ba2-b1 15. Kd1-e1 Bb1-a2 16. Ke1-f1 Ba2-b1 17. Kf1-g1 Bb1-a2 18. Kg1-h2 {
} Ba2-b1 19. Kh2-h1 Bb1-a2 20. Kh1-g1 Ba2-b1 21. Kg1-f1 Bb1-a2 22. Kf1-e1 Ba2-b1 23. Ke1-d1 Bb1-a2 {
} 24. Kd1-c1 h6-h5 25. Kc1-d1 Ba2-b1 26. Kd1-e1 Bb1-a2 27. Ke1-f1 Ba2-b1 28. Kf1-g1 Bb1-a2 29. Kg1-h2 {
} Ba2-b1 30. Kh2-h1Bb1-a2 31. Kh1-g1 Ba2-b1 32. Kg1-f1 Bb1-a2 33. Kf1-e1 Ba2-b1 34. Ke1-d1 Bb1-a2 {
} 35. Kd1-c1 f6-f5 36. Kc1-d1 Ba2-b1 37. Kd1-e1 Bb1-a2 38. Ke1-f1 Ba2-b1 39. Kf1-g1 Bb1-a2 40. Kg1-h2 {
} Ba2-b1 41. Kh2-h1 Bb1-a2 42. Kh1-g1 Ba2-b1 43. Kg1-f1 Bb1-a2 44. Kf1-e1 Ba2-b1 45. Ke1-d1 Bb1-a2 {
} 46. Kd1-c1 f5-f4 47. Kc1-d1 Ba2-b1 48. Kd1-e1 Bb1-a2 49. Ke1-f1 Ba2-b1 50. Kf1-g1 Bb1-a2 51. Kg1-h2 {
} Ba2-b1 52. Kh2-h1 Bb1-a2 53. Kh1-g1 Ba2-b1 54. Kg1-f1 Bb1-a2 55. Kf1-e1 Ba2-b1 56. Ke1-d1 Bb1-a2 {
} 57. Kd1-c1 f7-f6 58. Kc1-d1 Ba2-b1 59. Kd1-e1 Bb1-a2 60. Ke1-f1 Ba2-b1 61. Kf1-g1 Bb1-a2 62. Kg1-h2 {
} Ba2-b1 63. Kh2-h1 Bb1-a2 64. Kh1-g1 Ba2-b1 65. Kg1-f1 Bb1-a2 66. Kf1-e1 Ba2-b1 67. Ke1-d1 Bb1-a2 {
} 68. Kd1-c1 f6-f5 69. Kc1-d1 Ba2-b1 70. Kd1-e1 Bb1-a2 71. Ke1-f1 Ba2-b1 72. Kf1-g1 Bb1-a2 73. Kg1-h2 {
} Ba2-b1 74. Kh2-h1 Bb1-a2 75. Kh1-g1 Ba2-b1 76. Kg1-f1 Bb1-a2 77. Kf1-e1 Ba2-b1 78. Ke1-d1 Bb1-a2 {
} 79. Kd1-c1 c7-c6 80. Kc1-d1 Ba2-b1 81. Kd1-e1 Bb1-a2 82. Ke1-f1 Ba2-b1 83. Kf1-g1 Bb1-a2 84. Kg1-h2 {
} Ba2-b1 85. Kh2-h1 Bb1-a2 86. Kh1-g1 Ba2-b1 87. Kg1-f1 Bb1-a2 88. Kf1-e1 Ba2-b1 89. Ke1-d1 Bb1-a2 {
} 90. Kd1-c1 c6-c5 91. Kc1-d1 Ba2-b1 92. Kd1-e1 Bb1-a2 93. Ke1-f1 Ba2-b1 94. Kf1-g1 Bb1-a2 95. Kg1-h2 {
} Ba2-b1 96. Kh2-h1 Bb1-a2 97. Kh1-g1 Ba2-b1 98. Kg1-f1 Bb1-a2 99. Kf1-e1 Ba2-b1 100. Ke1-d1 Bb1-a2 {
} 101. Kd1-c1 Ba2*b3 102. Sa1*b3 c5-c4 103. Qc8*c4+ b7-b5 104. a5*b6ep.+
(C-)
3.Bxc7 Ka7 4.Bb6+ Ka6 5.Bd4 and Qc4+ soon…
Did I miss something?
Or even worse: 1. ba5 Ka6 2. Qd3 b5 3. ab6ep
In the intention, the triangulation is dualistic.
Is the following a possible fix?
EDIT: Oops, seems to lose one repetition, so it’s an ep93.
Bb8 can still be transferred to b6 and this frees Qc8 and white Knights. I do not see a clear sequence yet, but everything seems very suspicious.
After the black pawn moves on the king side are exhausted, how do you remove the bishop again from b6 after c5? Not saying that it’s not cooked, but I don’t see a clear way.
The following method seems to work.
1.ba Ka6 2.c8Q f3 3.Bc7
Black cannot move pawns as Bishop will return to b8.
So, 3…Ka7 4.Bb6+ Ka6 5.Qf5!
Black f-pawns are stopped. Now, white only need tempos to force h7-h6-h5.
When afterwards black play c5 white play Qxc5 f5 – Qc8 f4 – Bd4 f5 – Qc4+ having just enough black tempos in reserve.
So, the solution should at least be shortened by one more white king run by shifting pf6 to f5.
By the way, ‘C-‘ usually means ‘cooked’, ‘C?’ means ‘not computertested’.