Answers.Body: <p>White has made 6 captures. Clearly the h-pawn made at least 2, and the g-pawn 1. Meanwhile, the b-pawn made at least 1 (a7/c7), and the a-pawn at least 2 (to the b-file, then to either a7/c7). So White's c-d-e-f pawns made no captures, and thus Black's central pawn fence must have at some point had holes to allow them through. This requires 4 pawn captures, which is all that White has lost (knights and bishops).</p>
  
  <p>Here is a stupid method.</p>
  
  <pre><code>[fen ""]
  1. h4 Nh6 2. e4 f6 3. e5 Kf7 4. e6+ Kg6 5. f4 Ng4 6. f5+ Kh5 7. Nf3 a6 
  8. Ne5 fxe5 9. f6 Ra7 10. f7 Ra8 11. d4 Ra7 12. Bg5 Ra8 13. Bf6 exf6 
  14. Na3 Bb4+ 15. Ke2 Bd2 16. f8=R Bg5 17. hxg5+ Nh2 18. e7 Kh4 19. e8=R Ra7 
  20. Re6 Ra8 21. Rb6 Ra7 22. Rb3 Ra8 23. Re8 Ra7 24. Re6 Ra8 25. Reb6 Ra7 
  26. R6b4 Ra8 27. d5+ Kh5 28. g6 Rf8 29. Rg1 Rf7 30. gxf7 Ng4 31. f8=R Nh6 
  32. Re8 Nf7 33. g4+ Kh4 34. g5+ Kh5 35. Re6 Nc6 36. g6 Nb8 37. gxf7 Nc6 
  38. f8=R Nb8 39. Rc6 Ra7 40. Rcc3 Ra8 41. Re8 Ra7 42. Re6 Ra8 43. Rec6 Ra7 
  44. R6c4 Ra8 45. Rcg4 Ra7 46. R4g2 Ra8 47. Rbg4 Ra7 48. R4g3 Ra8 49. d6 c6 
  50. Nc4 Ra7 51. Ne3 Ra8 52. Nd5 cxd5 53. Rcf3 Kh6 54. c4 Ra7 55. c5 Ra8 
  56. c6 Ra7 57. c7 Ra8 58. Kf2 Ra7 59. Bb5 Ra8 60. Bc6 dxc6 61. Qe2 Bf5 
  62. c8=R Qe7 63. d7 Bd3 64. d8=R Bb5 65. a4 Ra7 66. axb5 Nd7 67. b6 Nc5 
  68. bxa7 Ne6 69. Rbe3 Nc7 70. b4 Qf7 71. a8=R Qg6 72. b5 Qg5 73. b6 Kg6 
  74. bxc7 Kf7 75. Ra7 Qf5 76. Rgc1 Qe6 77. Ke1 Qe7 78. Rd6 Qe6 79. Rcd8 Qe7 
  80. R8d7 Kf8 81. c8=R+ Qe8 82. Rcc7 Qf7 83. Re6 Qe7 84. Kd2 Qd6 85. Ree7 
  Qc5 86. Rf7+ Ke8 87. Rec3 Qb5 88. Qc4 Qb6 89. Qc5 Qb5 90. Qd6 Qb6 91. Kc2 
  Qc5 92. Kb2 Qc4 93. Rca3 Qc5 94. R3a2 Qb6+ 95. Rb3 Qb4 96. Rgc3 Qb5 97. Rc5 
  Qb6 98. Rb4 Qb5 99. Rgc2 Qxb4#
  </code></pre>
  
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2016-03-19T23:02:16.143Z
Answers.Text: Note that the fence closes halfway through, so the last four rooks must be inside of it.
Answers.UserDisplayName: Post-It-Note
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2016-03-19T06:48:52.193Z
Answers.LastActivityDate: 2016-03-19T07:02:54.133Z
Answers.LastEditDate: 2016-03-19T07:02:54.133Z
Answers.UserDisplayName: Post-It-Note
Body: <p>On his page on <a href="https://timkr.home.xs4all.nl/tour/horse.htm" rel="nofollow noreferrer">Harry Goldsteen's &quot;Horse Concoction,&quot;</a>, Tim Krabbé gives this position by &quot;N. Höeg, 1935.&quot;</p>
  <pre><code>[FEN &quot;4k3/1p4pp/p1pQ1p2/1q1pp3/8/8/1K6/8 b - - 0 1&quot;]
  [StartFlipped &quot;0&quot;]
  </code></pre>
  <p>With the stipulation &quot;add 10 white rooks; black mates in 1,&quot; he then claims that &quot;Retrograde-analysis will prove that these Rooks can only stand on a1, a2, a7, b4, c1, c2, c5, c7, d7 and f7, allowing Black to mate with Qxb4.&quot; This is the claimed solution,</p>
  <pre><code>[FEN &quot;4k3/RpRR1Rpp/p1pQ1p2/1qRpp3/1R6/8/RKR5/R1R5 b - - 0 1&quot;]
  [StartFlipped &quot;0&quot;]
  </code></pre>
  <p>However, I don't see how the supposed solution position could be legal. All of White's pawns have promoted, but because all of Black's pawns are present they must all have captured at least once to do so. But Black has only 5 missing pieces.</p>
  
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2016-03-19T06:10:49.887Z
Comments.Text: "All of White's pawns have promoted, but because all of Black's pawns are present they must all have captured at least once to do so." Das ist falsch. Black could also make pawn captures, allowing a free file. I haven't worked out yet if it is OK.
Comments.UserDisplayName: Post-It-Note
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2016-03-19T06:13:31.060Z
Comments.Text: Yes, the whole point is that four of the rooks must be promoted after the black pawn chain became reintact, while the other four came before/during the stage when there was access allowed. Which exactly accounts on the 7th rank holes (else Black is in check).
Comments.UserDisplayName: Post-It-Note
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2016-03-19T06:32:49.487Z
Comments.Text: Your diagram is wrong, the b1 rook should be on c5.
Comments.UserDisplayName: Post-It-Note
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2016-03-19T09:42:55.563Z
Comments.Text: Also, Black has 6 missing pieces, not 5 (both rooks,knights, bishops).
Comments.UserDisplayName: Post-It-Note
Tags: <problems><retrograde-analysis>
Title: How is this retrograde analysis position legal?
ThreadText: # How is this retrograde analysis position legal?
  
  (Asked by: username_0 on 2016-03-19)
  
  On his page on Harry Goldsteen's "Horse Concoction,", Tim Krabbé gives this position by "N. Höeg, 1935."
  [FEN "4k3/1p4pp/p1pQ1p2/1q1pp3/8/8/1K6/8 b - - 0 1"]
  [StartFlipped "0"]
  
  With the stipulation "add 10 white rooks; black mates in 1," he then claims that "Retrograde-analysis will prove that these Rooks can only stand on a1, a2, a7, b4, c1, c2, c5, c7, d7 and f7, allowing Black to mate with Qxb4." This is the claimed solution,
  [FEN "4k3/RpRR1Rpp/p1pQ1p2/1qRpp3/1R6/8/RKR5/R1R5 b - - 0 1"]
  [StartFlipped "0"]
  
  However, I don't see how the supposed solution position could be legal. All of White's pawns have promoted, but because all of Black's pawns are present they must all have captured at least once to do so. But Black has only 5 missing pieces.
  
  **Tags:** problems, retrograde-analysis
  
  > **Comments:**
  > **username_1**: "All of White's pawns have promoted, but because all of Black's pawns are present they must all have captured at least once to do so." Das ist falsch. Black could also make pawn captures, allowing a free file. I haven't worked out yet if it is OK.  
  > **username_1**: Yes, the whole point is that four of the rooks must be promoted after the black pawn chain became reintact, while the other four came before/during the stage when there was access allowed. Which exactly accounts on the 7th rank holes (else Black is in check).  
  > **username_1**: Your diagram is wrong, the b1 rook should be on c5.  
  > **username_1**: Also, Black has 6 missing pieces, not 5 (both rooks,knights, bishops).  
  
  ---
  
  ## 1 answers
  
  ### Answer by username_1 on 2016-03-19. Score: 17 | Accepted answer
  
  White has made 6 captures. Clearly the h-pawn made at least 2, and the g-pawn 1. Meanwhile, the b-pawn made at least 1 (a7/c7), and the a-pawn at least 2 (to the b-file, then to either a7/c7). So White's c-d-e-f pawns made no captures, and thus Black's central pawn fence must have at some point had holes to allow them through. This requires 4 pawn captures, which is all that White has lost (knights and bishops).
  Here is a stupid method.
  [fen ""]
  1. h4 Nh6 2. e4 f6 3. e5 Kf7 4. e6+ Kg6 5. f4 Ng4 6. f5+ Kh5 7. Nf3 a6 
  8. Ne5 fxe5 9. f6 Ra7 10. f7 Ra8 11. d4 Ra7 12. Bg5 Ra8 13. Bf6 exf6 
  14. Na3 Bb4+ 15. Ke2 Bd2 16. f8=R Bg5 17. hxg5+ Nh2 18. e7 Kh4 19. e8=R Ra7 
  20. Re6 Ra8 21. Rb6 Ra7 22. Rb3 Ra8 23. Re8 Ra7 24. Re6 Ra8 25. Reb6 Ra7 
  26. R6b4 Ra8 27. d5+ Kh5 28. g6 Rf8 29. Rg1 Rf7 30. gxf7 Ng4 31. f8=R Nh6 
  32. Re8 Nf7 33. g4+ Kh4 34. g5+ Kh5 35. Re6 Nc6 36. g6 Nb8 37. gxf7 Nc6 
  38. f8=R Nb8 39. Rc6 Ra7 40. Rcc3 Ra8 41. Re8 Ra7 42. Re6 Ra8 43. Rec6 Ra7 
  44. R6c4 Ra8 45. Rcg4 Ra7 46. R4g2 Ra8 47. Rbg4 Ra7 48. R4g3 Ra8 49. d6 c6 
  50. Nc4 Ra7 51. Ne3 Ra8 52. Nd5 cxd5 53. Rcf3 Kh6 54. c4 Ra7 55. c5 Ra8 
  56. c6 Ra7 57. c7 Ra8 58. Kf2 Ra7 59. Bb5 Ra8 60. Bc6 dxc6 61. Qe2 Bf5 
  62. c8=R Qe7 63. d7 Bd3 64. d8=R Bb5 65. a4 Ra7 66. axb5 Nd7 67. b6 Nc5 
  68. bxa7 Ne6 69. Rbe3 Nc7 70. b4 Qf7 71. a8=R Qg6 72. b5 Qg5 73. b6 Kg6 
  74. bxc7 Kf7 75. Ra7 Qf5 76. Rgc1 Qe6 77. Ke1 Qe7 78. Rd6 Qe6 79. Rcd8 Qe7 
  80. R8d7 Kf8 81. c8=R+ Qe8 82. Rcc7 Qf7 83. Re6 Qe7 84. Kd2 Qd6 85. Ree7 
  Qc5 86. Rf7+ Ke8 87. Rec3 Qb5 88. Qc4 Qb6 89. Qc5 Qb5 90. Qd6 Qb6 91. Kc2 
  Qc5 92. Kb2 Qc4 93. Rca3 Qc5 94. R3a2 Qb6+ 95. Rb3 Qb4 96. Rgc3 Qb5 97. Rc5 
  Qb6 98. Rb4 Qb5 99. Rgc2 Qxb4#
  
  > **Comments:**
  > **username_1**: Note that the fence closes halfway through, so the last four rooks must be inside of it.  
  
  ---
  
  