Least number of moves to achieve a complete switchover of nonpawns in starting setup

Question

 [fen "RNBQKBNR/8/8/pppppppp/PPPPPPPP/8/8/rnbqkbnr w - - 0 1"]

During a comp, my friend an I had a team kill and we decided to make this position(legally) before drawing. Then we continued to find more and more optimal ways of getting the position until I found this:

[FEN ""]

1. a4 a5 2. Ra3 Ra6 3. Rb3 Rc6 4. Rb6 Rc3 5. Ra6 Ra3 6. Ra8 Ra1 7. h4 h5 8. Rh3 Rh6 9. Rg3 Rf6 10. Rg6 Rf3 11. Rh6 Rh3 12. Rh8 Rh1 13. e3 e6 14. Qg4 Ke7 15. Ke2 Kd6 16. Kf3 Qf6+ 17. Ke4 Kc5 18. d3 Kb4 19. c3+ Kb3 20. Qg5 Kc2 21. Na3+ Kd1 22. Nh3 Qf3+ 23. Kd4 Ke1 24. Qd8 Qd1 25. Ke5 Nh6 26. Ng5 g6 27. Kf6 Bg7+ 28. Ke7 Nc6+ 29. Ke8 Nd4 30. c4 Nf3 31. b3 b6 32. Nb5 Ng1 33. Ba3 Bb2 34. Bf8 Bc1 35. g3 Bb7 36. Bh3 d6 37. Nh7 e5 38. Bc8 Bg2 39. Nf6 Bf1 40. Ng8 Nf5 41. Na7 Nd4 42. Nc6 Ndf3 43. Nb8 Nd2 44. b4 Nb1 45. d4 d5 46. e4 c5 47. f3 b5 48. f4 f5 49. g4 g5

Is there any faster way of achieving this position? And (as an extra question) does anyone have a position that have the longest legal optimal series of moves to achieve it?

Answer 1

As @chakerian's calculations show, 40 moves is the minimum. After a bit of puzzling, I found the solution.

[FEN ""]

1. a4 {First, we need to get the rooks in position. They'll be hard to swap once the board is full.} a5 2. h4 h5 3. Rh3 Rh6 4. Rg3 Rf6 5. Rg6 Rf3 6. Rh6 Rh3 7. Ra3 Ra6 8. Rc3 Rb6 9. Rcc6 Rbb3 10. Ra6 Ra3 {Now, we can start with the knights. They're tricky because they need to swap places via the four center squares. Note that we cannot move the center pawns *before* the knights.} 11. Nf3 Nc6 12. Ne5 Nd4 13. Nc6 Nf6 14. Nc3 Nd5 15. Ne4 Nc3 {It's time for the kings to come out of their hiding, but they have do this carefully because of the knights.} 16. Rh8 d5 17. Ra8 Kd7 18. Nf6+ Kd6 19. e4 Ra1 20. Ba6 b5 21. Nb8 Kc5 {The black monarch chooses queenside - that means his colleague has to take the other one.} 22. Ng8 Nde2 23. d4+ Kc4 24. Kd2 Rh1 25. Ke3 Ng1 26. Kf4 Bh3 27. g4 f5 28. Kg5 e5+ 29. Kg6 Bf1 30. Bh6 {Note how the c1-h6 and a3-f8 diagonals have been left open for the transfer of the last four pieces.} Qg5+ 31. Kf7 Ba3 32. Ke8 Qc1 33. Qd2 Nb1 34. Qg5 {OK, we're almost there now, from here it's a walk in the park.} Qd1 35. b4 Kc3 36. Qd8 g5 37. Bf8 c5 38. Bc8 Kd2 39. c4 Bc1 40. f4 Ke1 {And we're done!}

To answer the second question: you're looking for the longest unique proof game. I found the longest known one here and here.

[Title "Dmitri W. Pronkin & Andriy Frolkin, Die Schwalbe 06/1989, 1st Prize, PG in 57.5"]
[FEN ""]

1. a4 h5 2. a5 h4 3. a6 h3 4. axb7 hxg2 5. h4 d5 6. h5 d4 7. h6 d3 8. h7 dxc2 9. d4 a5 10. Bh6 c1=R 11. e4 Rc5 12. Ne2 Rh5 13. e5 c5 14. e6 Nc6 15. b8=R a4 16. Rb4 a3 17. Ra4 c4 18. b4 c3 19. b5 c2 20. b6 c1=R 21. b7 Rc4 22. b8=R Qa5+ 23. Rbb4 Bb7 24. N1c3 O-O-O 25. exf7 e5 26. Rc1 Bc5 27. f8=R a2 28. Rf3 a1=R 29. Na2 g1=R 30. Rfa3 Rg6 31. f4 Re6 32. f5 g5 33. f6 g4 34. f7 g3 35. f8=R g2 36. Rf5 g1=R 37. Bf8 Rg7 38. Ng3 e4 39. Bd3 e3 40. O-O e2 41. Rcc3 e1=R 42. Bc2 R1e3 43. d5 Rdd7 44. d6 Rdf7 45. d7+ Kb8 46. Qd6+ Ka8 47. Qc7 Nge7 48. d8=R+ Nc8 49. Rdd3 Rhg8 50. h8=R Rae1 51. Rh6 R1e2 52. R1f2 Rce4 53. Kf1 Bd4 54. Rfc5 Ne5 55. Nf5 Nc4 56. Nd6 Nb2 57. Rbc4 Nb6 58. Qb8+

What this really means, is that the final position can be reached in 57.5 moves (not less), and that there is only one way to do this. In your example, moves 48 and 49 (and many others) could be switched. Also, currently there is no other position with this characteristic known which takes more moves to obtain.

Answer 2

Shouldn't the minimum be 40?. I don't have any result to show yet but it seems that:

Pawns (black and white): 8 moves

Rooks (black and white): 10 moves

Kings (black and white): 7 moves

Queens (black and white): 3 moves

Knights (black and white): 8 moves

Bishops (black and white): 4 moves

I could be wrong

Something to consider is that the e pawns for black and white can't be pushed up together until the kings have passed to their opposite positions.

Possibly it takes an extra two moves for the last two knights in which case it would be 42.

Answer 3

I came up with this, which is 43 moves, but I think it could be optimized further:

[fen ""]
1. h4 h5 2. Rh3 Rh6 3. Rg3 Rf6 4. Rg6 Rf3
5. Rh6 Rh3 6. Rh8 Rh1 7. a4 a5 8. Ra3 Ra6
9. Rc3 Rb6 10. Rc6 Rb3 11. Ra6 Ra3 12. Ra8 Ra1
13. e4 d5 14. Ke2 Qd7 15. Ke3 Qg4 16. Qf3 Kd7
17. Qf4 Kc6 18. Kd4 Qd1 19. Ke5 Bh3 20. g4 f5
21. Ke6 Kc5 22. Kf7 Kd4 23. Ba6 b5 24. Bc8 Bf1
25. c4 Kd3 26. Nf3 Ke2 27. Ne5 Nc6 28. Nd7 Ke1
29. Ke8 Nd4 30. Qd6 Nf3 31. Nb8 Ng1 32. Qd8 e5
33. d4 Ba3 34. b4 Ne7 35. Bh6 g5 36. Bf8 Bc1
37. Nd2 Ng6 38. Nb3 Nf4 39. Nc5 Nfe2 40. Ncd7 Nc3
41. Nf6 Nb1 42. Ng8 c5 43. f4

EDIT: Okay, I think this is the optimized one:

[fen ""]
1. h4 a5 2. a4 h5 3. Rh3 Ra6 4. Raa3 Rhh6
5. Rhf3 Rac6 6. Rab3 Rhg6 7. Rf6 Rc3 8. Rbb6 Rgg3
9. Rh6 Ra3 10. Rh8 Ra1 11. Ra6 Rh3 12. Ra8 Rh1
13. e4 d5 14. Ke2 Bh3 15. Ke3 Kd7 16. Kf4 Kc6
17. Qg4 Kc5 18. Qc8 Qd7 19. Ke5 Qg4 20. Qd8 f5
21. Ke6 Qd1 22. Kf7 Kd4 23. Ba6 Nc6 24. c4 Kd3
25. Nf3 Ke2 26. Nd4+ Ke1 27. g4 Ne5+ 28. Ke8 Nf3
29. Nc6 Ng1 30. Nb8 Bf1 31. d4 b5 32. Bh6 e5
33. Bc8 Ba3 34. b4 g5 35. Bf8 Bc1 36. Nd2 Ne7
37. Nb3 Ng6 38. Nc5 Nf4 39. Ncd7 Nfe2 40. Nf6 Nc3
41. Ng8 Nb1 42. f4 c5

Answer 4

Here is a small improvement. It gains a tempo by avoiding Ke4-d4-d5.

[FEN ""] 

1. a4 a5 2. Ra3 Ra6 3. Rb3 Rc6 4. Rb6 Rc3 5. Ra6 Ra3 6. Ra8 Ra1 7. h4 h5 8. Rh3 
Rh6 9. Rg3 Rf6 10. Rg6 Rf3 11. Rh6 Rh3 12. Rh8 Rh1 13. e3 e6 14. Qg4 Ke7 15. Ke2 
Kd6 16. Kf3 Qf6+ 17. Ke4 Kc5 18. d3 Kb4 19. c3+ Kb3 20. Qg5 Kc2 21. Na3+ Kd1 22. 
Nh3 Qf3+ 23. Ke5 Ke1 24. Qd8 Qd1 25. Ng5 Nh6 26. c4 g6 27. Kf6 Bg7+ 28. Ke7 Nc6+ 
29. Ke8 Nd4 30. b3 Nf3 31. Nb5 Ng1 32. Ba3 Bb2 33. Bf8 Bc1 34. g3 b6 35. Bh3 Bb7 
36. Nh7 Bg2 37. Nf6 d5 38. Ng8 e5 39. Bc8 Bf1 40. Na7 Nf5 41. Nc6 Nd4 42. Nb8 
Ndf3 43. b4 Nd2 44. d4 Nb1 45. e4 b5 46. g4 g5 47. f3 c5 48. f4 f5 *

Note that the absolute minimum amount of moves to achieve the position is at least 38, and without 'crossing the knights' and with moving the bishops in 3 turns the minimum is already 44.

So that would mean that without drastic changes, and improvement would need to depend on only 3 additional 'suboptimal' moves, like moving a pawn twice. As such I don't think there is much room left for improvement, but perhaps 1 or 2 moves could still be shaved off.

Answer 5

I interpret the OP's <a position that have the longest legal optimal series of moves to achieve it?> as a request for a position which, for the largest known n, can be reached in n moves but cannot be reached in fewer. I don't see any request for there to be a unique n-move sequence.

[Title "Remotest possible chess position?"]
[FEN "Kr6/8/1r6/8/8/1PPPPpPP/kpbrrP1R/qbrnrrbn w KQkq - 0 1"]
[StartFlipped "0"]

One of the shortest known series of moves that reach this position is:

[FEN ""]
[StartPly "370"]

1. a4 c5 2. a5 c4 3. a6 c3 4. b3 d5 5. Bb2 cxb2 6. Na3 Kd7 7. Nc4 Kc6 8. Ne5+ Kb5 9. Ra5+ Kb4 10. Rb5+ Ka3 11. Nef3 Ka2 12. c3 d4 13. Qc1 e5 14. Kd1 e4 15. Kc2 e3 16. Kd3 g5 17. Ke4 g4 18. Kf4 d3 19. Qc2 dxc2 20. h3 c1=R 21. Rh2 Ra1 22. Rh1 Kb1 23. Rh2 Kc1 24. Rh1 Bf5 25. Rh2 Bb1 26. d3 f5 27. Nd2 exd2 28. Nf3 Nf6 29. Kg5 Bd6 30. Kh6 Nd5 31. Kg7 Ne3 32. Kf7 h5 33. Kg7 Ra4 34. Kf7 Ba2 35. Rb6 d1=B 36. Kg7 Bc2 37. Kf7 Bcb1 38. Kg7 Nd1 39. Kf7 g3 40. Kg7 gxh2 41. e3 h1=R 42. Be2 Nd7 43. Nh2 Nc5 44. Bg4 Ne4 45. Kf7 hxg4 46. Kg7 g3 47. Kf7 gxh2 48. Kg7 Re1 49. Kf7 Re2 50. Kg7 Rc2 51. Kf7 h1=R 52. Kg7 Re1 53. Kf7 Bh2 54. Kg7 Ng3 55. Kf7 Nh1 56. Kg7 Rah4 57. Kf7 Kd2 58. Kg7 Ke2 59. Kf7 Rc1 60. Kg7 Bc2 61. Kf7 Ra1 62. Kg7 Bab1 63. Kf7 Ra5 64. Kg7 Ba2 65. g3 Kf3 66. Rb4 Rb5 67. Kf7 Re2 68. Kg7 Bcb1 69. Kf7 Rc2 70. Kg7 Rc1 71. Kf7 Bc2 72. Kg7 Ra1 73. Kf7 Bab1 74. Kg7 Ra5 75. Ra4 bxa6 76. Ra1 Ba2 77. Rc1 Bcb1 78. Rc2 f4 79. Re2 Bc2 80. Re1 Bab1 81. Rg1 Ra1 82. Rg2 Ba2 83. Kf7 Rc1 84. Kg7 Bcb1 85. Kf7 Rc2 86. Kg7 Re2 87. Kf7 Re1 88. Kg7 Rg1 89. Kf7 Bc2 90. Kg7 Bab1 91. Kf7 Ra5 92. Kg7 Ra1 93. Kf7 Ba2 94. Kg7 Rc1 95. Kf7 Bcb1 96. Kg7 Rc2 97. Kf7 Re2 98. Kg7 Ree1 99. Kf7 Ref1 100. Kg7 Bc2 101. Kf7 Bab1 102. Kg7 a5 103. Kf7 a4 104. Kg7 a3 105. Kf7 a2 106. Kg7 a1=R 107. Kf7 Ba2 108. Kg7 Rc1 109. Kf7 Bcb1 110. Kg7 Rc2 111. Kf7 Re2 112. Kg7 Ree1 113. Kf7 Bc2 114. Kg7 Bab1 115. Kf7 a5 116. Kg7 a4 117. Kf7 a3 118. Kg7 a2 119. Kf7 a1=R 120. Kg7 Ba2 121. Kf7 Rc1 122. Kg7 Bcb1 123. Kf7 Rc2 124. Kg7 Ke2 125. Kf7 Kd2 126. Kg7 Kc1 127. Kf7 Rce2 128. Kg7 Kd2 129. Kf7 Bc2 130. Kg7 Bab1 131. Kf7 Ra1 132. Kg7 Ba2 133. Kf7 Rc1 134. Kg7 Bcb1 135. Kf7 Rc2 136. Kg7 Kc1 137. Kf7 Rcd2 138. Kg7 Bc2 139. Kf7 Kb1 140. Kg7 Ka1 141. Kf7 Bab1 142. Kg7 Ka2 143. Kf7 Ka3 144. Kg7 Ba2 145. Kf7 Bcb1 146. Kg7 Rc2 147. Kf7 Rc1 148. Kg7 Bc2 149. Kf7 Ra1 150. Kg7 Bcb1 151. Kf7 Rc2 152. Kg7 Qd5 153. Kf6 Qf3 154. Kf7 Qe2 155. Kg7 Qd2 156. Kf7 Qc1 157. Kg7 Rce2 158. Kf7 Qd2 159. Kg7 Bc2 160. Kf7 Rc1 161. Kg7 Bcb1 162. Kf7 Rc2 163. Kg7 Qc1 164. Kf7 Rcd2 165. Kg7 f3 166. Kf7 Bc2 167. Kg7 Qa1 168. Kf7 Bcb1 169. Kg7 Rc2 170. Kf7 Rc1 171. Kg7 Bc2 172. Kf7 Bab1 173. Kg7 Ka2 174. Kf7 Red2 175. Ke7 Ree2 176. Kd7 Rfe1 177. Kc7 Rgf1 178. Kb7 Bg1 179. Ka7 R4h6 180. Rh2 Rb6 181. Rg2 Rh7+ 182. Ka8 Rhb7 183. Rh2 Rb8+

The construction was published in Am Rande des Schachbretts in 1947 by Luigi Ceriani and Karl Fabel. Here is a direct link to its entry in PDB.