Answers.Body: <p>I got this, which is much shorter.</p>
  <pre><code>[FEN &quot;&quot;]
  
  1. a4 a5 2. b4 b5 3. c4 c5 4. d4 d5 5. e4 e5 6. f4 f5 7. g4 g5 8. h4 h5 9. bxa5 Bd6 10. axb5 Be6 11. dxc5 Nf6 12. cxd5 O-O 13. fxe5 Rf7 14. exf5 Ra6 15. gxh5 Rb7 16. hxg5 Rbb6 17. bxa6 Nc6 18. axb6 Qe7 19. dxc6 Kh8 20. cxd6 Ng8 21. a7 Qf6 22. gxf6 Ne7 23. b7 Bd7 24. e6 Ng8 25. c7 Kh7 26. f7 Kg7 27. e7 Kf6 28. h6 Ke5 29. f6 Kf5 30. a8=Q Kg6 31. b8=Q Ba4 32. c8=Q Bb5 33. f8=Q Bc4 34. d7 Bd5 35. d8=Q Be4 36. f7 Bf5 37. e8=Q Bd7 38. h7 Bc6 39. h8=Q Bb7 40. Qcxb7 Kf5 41. fxg8=Q#
  </code></pre>
  <p>It's one move above the lower bound, but I feel that this is optimal.</p>
  <hr />
  <p>As @Evargalo suggested, here is @user1583209's proof of optimality from their comments.</p>
  <p>&quot;I have a proof now: Start by labeling the columns/files by numbers, i.e. a=1, b=2...h=8. Then for each pawn calculate the sum of the respective files. So in the initial position this sum is 1+2+3...+8=36. Now, if a pawn captures, it changes its file by &quot;1&quot; (either plus or minus 1). In order to achieve a solution in 40 moves, we have to queen the pawns on each of the squares on the 8th rank. This means that in the final position the sum of files (for pawns+promoted pawns now) needs to be 36 again. However we also need to do 15 captures with these pawns.
  Each capture changing the file sum by 1, it is impossible to end up with 36 as sum, basically because 15 is an odd number.  (If you have x captures to the left and y captures to the right you end up with equations: x-y=0 and x+y=15 which has no integer solution) That proves that it is impossible to achieve a solution in 40 moves and the solution with 41 moves given by orlp is optimal.&quot;</p>
  
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-02-17T00:56:53.940Z
Answers.Text: After some thinking I'm more or less convinced that the 40 move solution cannot exist; that would require that at one point 7 white queens would be standing on the 8th rank while there would only be one uncaptured black piece besides the king itself (say, after move 39 by white). But this would be a contradiction since the black king would then lack shelter anywhere on the board. So I think that this solution is at most 2 moves from an optimal one.
Answers.UserDisplayName: Scounged
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-02-17T01:09:31.640Z
Answers.Text: @Scounged It is possible to set up a position with 7 white queens and 1 black piece  on the 8th rank, one white pawn on the 7th rank and the king not in check. For instance: black king on g6, white pawn on f7, black knight on g8, white queens on all remaining squares on the 8th rank. I am not claiming it is possible to achieve this setup, just saying that more careful analysis is needed..
Answers.UserDisplayName: user1583209
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-02-17T01:16:32.377Z
Answers.Text: @Scounged I shaved off one more move and got 42.
Answers.UserDisplayName: orlp
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-02-17T01:23:17.453Z
Answers.Text: @user1583209 Indeed, that looks possible, but it's not clear how it would arise either. For instance, what are white's and black's last moves before such a position? But as you noted, there are some special cases that need to be investigated before I can be certain that a 40 move solution cannot be achieved, and if one is incredibly lucky it could be possible to retrace the game backwards from one such special case position to find a 40 move solution. Are there any other 'safe squares' than b6 and g6 for the black king that need to be investigated? I don't see any.
Answers.UserDisplayName: Scounged
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-02-17T01:28:52.230Z
Answers.Text: @Scounged Shaved off another with user1583209's idea, now 1 above the lower bound.
Answers.UserDisplayName: orlp
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-02-17T01:30:38.420Z
Answers.Text: @user1583209 In your setup you have exactly one column with doubled pawns. As you can see in my solution, we are left with one major piece, which can not be captured without doubling another column (or tripling). That makes me believe my solution is optimal (although it's not a hard proof).
Answers.UserDisplayName: orlp
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-02-17T02:07:44.090Z
Answers.Text: @Scounged *Are there any other 'safe squares' than b6 and g6 for the black king that need to be investigated? I don't see any.* I don't see any either. Only b6 and g6 are safe and only in a setup with a white pawn on c7 (f7) and a black piece on b8 (g8).
Answers.UserDisplayName: user1583209
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-02-17T02:19:37.497Z
Answers.Text: @orlp Good point. So it comes down to the question, whether with 15 pawn captures you can "undouble"/"untriple" all pawns.
Answers.UserDisplayName: user1583209
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-02-17T02:25:25.390Z
Answers.Text: @orlp I would guess you need an even number of captures to do that but have no prroof.
Answers.UserDisplayName: user1583209
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-02-17T03:19:21.160Z
Answers.Text: I have a proof now: Start by labeling the columns/ranks by numbers, i.e. a=1, b=2...h=8. Then for each pawn calculate the sum of the respective ranks. So in the initial position this sum is 1+2+3...+8=36. Now, if a pawn captures it changes its rank by "1" (either plus or minus 1). In order to achieve a solution in 40 moves, we have to queen the pawns on each of the squares on the 8th row. This means that in the final position the sum of ranks (for pawns+promoted pawns now) needs to be 36 again. However we also need to do 15 captures with these pawns.
Answers.UserDisplayName: user1583209
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-02-17T03:20:58.153Z
Answers.Text: Each capture changing the rank sum by 1, it is impossible to end up with 36 as sum, basically because 15 is an odd number.  (If you have x captures to the left and y captures to the right you end up with equations: x-y=0 and x+y=15 which has no integer solution) That proves that it is impossible to achieve a solution in 40 moves and the solution with 41 moves given by orlp is optimal.
Answers.UserDisplayName: user1583209
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-02-17T09:09:08.167Z
Answers.Text: @user1583209 Neat parity argument, but you mean files, not ranks.
Answers.UserDisplayName: Rosie F
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-02-17T11:04:16.880Z
Answers.Text: @user1583209 That's exactly what I meant with my comment, nicely formalized.
Answers.UserDisplayName: orlp
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2018-06-08T13:21:03.937Z
Answers.Text: I think @user1583209 's nice argument should be moved from the comments into the body of the answer, so that it doesn't get lost.
Answers.UserDisplayName: Evargalo
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2018-02-16T23:53:08.943Z
Answers.LastActivityDate: 2020-12-14T23:55:45.810Z
Answers.LastEditDate: 2020-12-14T23:55:45.810Z
Answers.UserDisplayName: orlp
Body: <p>What is the shortest game with the largest material imbalance? This of course means that one side will have 9 queens, 2 rooks, 2 bishops, 2 knights, and a king, and a bare king on the other side. From some experimenting, I got this:</p>
  
  <pre><code>[FEN ""]
  
  1.h4 g5 2.hxg5 h6 3.gxh6 Rh7 4.g4 Rg7 5.h7 Rg6 6.h8=Q Bh6 7.g5 f5 8.f4 e6 9.e4 d5 10.d4 c5 11.c4 b5 12.b4 a5 13.a4 Qe7 14.exf5 Bb7 15.fxe6 Bc6 16.cxd5 Bb7 17.bxc5 Bc8 18.axb5 Bb7 19.c6 Bc8 20.Rxa5 Ra7 21.b6 Rd7 22.Qe5 Nf6 23.b7 Kf8 24.c7 Rd6 25.f5 Kg8 26.Nc3 Bd7 27.c8=Q+ Kh7 28.Qc4 Nc6 29.b8=Q Be8 30.Bd2 Qf8 31.e7 Qg7 32.Bg2 Bf7 33.e8=Q Re6 34.dxe6 Bg8 35.d5 Nb4 36.d6 Nc6 37.d7 Nb4 38.d8=Q Nc6 39.e7 Nb4 40.Qdd7 Na6 41.Qed8 Nc5 42.e8=Q Nd3+ 43.Qcxd3 Ng4 44.f6 Ne3 45.f7 Nf5 46.f8=Q Qf7 47.Qfb4 Rg7 48.g6+ Kh8 49.gxf7 Ne7 50.Q3d6 Kh7 51.Q6xe7 Kh8 52.f8=Q Kh7 53.Qed5 Rf7 54.Bh3 Kh8 55.Q7xf7 Bg7 56.Ke2 Kh7 57.Ke3 Kh6 58.Kf4 Kh7 59.Q8xg8+ Kh6 60.Qgxg7#  0-1
  </code></pre>
  
  <p>This can of course be improved though, as I was essentially moving pieces around randomly.</p>
  
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2018-02-16T23:00:51.417Z
Comments.Text: Well, 48 moves is the absolute theoretical minimum. What needs to be done is to see whether it's possible to get this material imbalance with white only making pawn moves.
Comments.UserDisplayName: Scounged
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2018-02-16T23:10:44.193Z
Comments.Text: @Scounged Problem is taking all the pieces right away leaves the enemy king vulnerable to stalemate or checkmate.
Comments.UserDisplayName: ericw31415
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2018-02-16T23:15:32.583Z
Comments.Text: @Scounged How do you come up with the number 48?
Comments.UserDisplayName: user1583209
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2018-02-16T23:37:51.527Z
Comments.Text: @user1583209 Sorry, I was probably being unclear, we have a lower theoretical limit in that white has to make *at least* 48 (8*6) moves (since all eight white pawns need to promote, and thus each white pawn needs to make at least 6 moves). I should have made it more clear that this is only a lower limit, and that it may not be possible to achieve this lower limit.
Comments.UserDisplayName: Scounged
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2018-02-17T00:43:15.230Z
Comments.Text: Ok, so I just miscalculated totally and forgot that the white pawns begin on the second rank. Of course the lower limit is 40 moves (8*5 pawn moves).
Comments.UserDisplayName: Scounged
Tags: <problems><records><material-balance>
Title: Shortest sequence of moves to create biggest material imbalance
ThreadText: # Shortest sequence of moves to create biggest material imbalance
  
  (Asked by: username_0 on 2018-02-16)
  
  What is the shortest game with the largest material imbalance? This of course means that one side will have 9 queens, 2 rooks, 2 bishops, 2 knights, and a king, and a bare king on the other side. From some experimenting, I got this:
  
  ```
  [FEN ""]
  
  1.h4 g5 2.hxg5 h6 3.gxh6 Rh7 4.g4 Rg7 5.h7 Rg6 6.h8=Q Bh6 7.g5 f5 8.f4 e6 9.e4 d5 10.d4 c5 11.c4 b5 12.b4 a5 13.a4 Qe7 14.exf5 Bb7 15.fxe6 Bc6 16.cxd5 Bb7 17.bxc5 Bc8 18.axb5 Bb7 19.c6 Bc8 20.Rxa5 Ra7 21.b6 Rd7 22.Qe5 Nf6 23.b7 Kf8 24.c7 Rd6 25.f5 Kg8 26.Nc3 Bd7 27.c8=Q+ Kh7 28.Qc4 Nc6 29.b8=Q Be8 30.Bd2 Qf8 31.e7 Qg7 32.Bg2 Bf7 33.e8=Q Re6 34.dxe6 Bg8 35.d5 Nb4 36.d6 Nc6 37.d7 Nb4 38.d8=Q Nc6 39.e7 Nb4 40.Qdd7 Na6 41.Qed8 Nc5 42.e8=Q Nd3+ 43.Qcxd3 Ng4 44.f6 Ne3 45.f7 Nf5 46.f8=Q Qf7 47.Qfb4 Rg7 48.g6+ Kh8 49.gxf7 Ne7 50.Q3d6 Kh7 51.Q6xe7 Kh8 52.f8=Q Kh7 53.Qed5 Rf7 54.Bh3 Kh8 55.Q7xf7 Bg7 56.Ke2 Kh7 57.Ke3 Kh6 58.Kf4 Kh7 59.Q8xg8+ Kh6 60.Qgxg7#  0-1
  
  ```
  
  This can of course be improved though, as I was essentially moving pieces around randomly.
  
  **Tags:** problems, records, material-balance
  
  > **Comments:**
  > **username_1**: Well, 48 moves is the absolute theoretical minimum. What needs to be done is to see whether it's possible to get this material imbalance with white only making pawn moves.  
  > **username_0**: @username_1 Problem is taking all the pieces right away leaves the enemy king vulnerable to stalemate or checkmate.  
  > **username_2**: @username_1 How do you come up with the number 48?  
  > **username_1**: @username_2 Sorry, I was probably being unclear, we have a lower theoretical limit in that white has to make *at least* 48 (8*6) moves (since all eight white pawns need to promote, and thus each white pawn needs to make at least 6 moves). I should have made it more clear that this is only a lower limit, and that it may not be possible to achieve this lower limit.  
  > **username_1**: Ok, so I just miscalculated totally and forgot that the white pawns begin on the second rank. Of course the lower limit is 40 moves (8*5 pawn moves).  
  
  ---
  
  ## 1 answers
  
  ### Answer by username_3 on 2018-02-16. Score: 12 | Accepted answer
  
  I got this, which is much shorter.
  
  ```
  [FEN ""]
  
  1. a4 a5 2. b4 b5 3. c4 c5 4. d4 d5 5. e4 e5 6. f4 f5 7. g4 g5 8. h4 h5 9. bxa5 Bd6 10. axb5 Be6 11. dxc5 Nf6 12. cxd5 O-O 13. fxe5 Rf7 14. exf5 Ra6 15. gxh5 Rb7 16. hxg5 Rbb6 17. bxa6 Nc6 18. axb6 Qe7 19. dxc6 Kh8 20. cxd6 Ng8 21. a7 Qf6 22. gxf6 Ne7 23. b7 Bd7 24. e6 Ng8 25. c7 Kh7 26. f7 Kg7 27. e7 Kf6 28. h6 Ke5 29. f6 Kf5 30. a8=Q Kg6 31. b8=Q Ba4 32. c8=Q Bb5 33. f8=Q Bc4 34. d7 Bd5 35. d8=Q Be4 36. f7 Bf5 37. e8=Q Bd7 38. h7 Bc6 39. h8=Q Bb7 40. Qcxb7 Kf5 41. fxg8=Q#
  
  ```
  
  It's one move above the lower bound, but I feel that this is optimal.
  
  ---
  
  As @username_5 suggested, here is @username_2's proof of optimality from their comments.
  
  "I have a proof now: Start by labeling the columns/files by numbers, i.e. a=1, b=2...h=8. Then for each pawn calculate the sum of the respective files. So in the initial position this sum is 1+2+3...+8=36. Now, if a pawn captures, it changes its file by "1" (either plus or minus 1). In order to achieve a solution in 40 moves, we have to queen the pawns on each of the squares on the 8th rank. This means that in the final position the sum of files (for pawns+promoted pawns now) needs to be 36 again. However we also need to do 15 captures with these pawns.
  Each capture changing the file sum by 1, it is impossible to end up with 36 as sum, basically because 15 is an odd number.  (If you have x captures to the left and y captures to the right you end up with equations: x-y=0 and x+y=15 which has no integer solution) That proves that it is impossible to achieve a solution in 40 moves and the solution with 41 moves given by username_3 is optimal."
  
  > **Comments:**
  > **username_1**: After some thinking I'm more or less convinced that the 40 move solution cannot exist; that would require that at one point 7 white queens would be standing on the 8th rank while there would only be one uncaptured black piece besides the king itself (say, after move 39 by white). But this would be a contradiction since the black king would then lack shelter anywhere on the board. So I think that this solution is at most 2 moves from an optimal one.  
  > **username_2**: @username_1 It is possible to set up a position with 7 white queens and 1 black piece  on the 8th rank, one white pawn on the 7th rank and the king not in check. For instance: black king on g6, white pawn on f7, black knight on g8, white queens on all remaining squares on the 8th rank. I am not claiming it is possible to achieve this setup, just saying that more careful analysis is needed..  
  > **username_3**: @username_1 I shaved off one more move and got 42.  
  > **username_1**: @username_2 Indeed, that looks possible, but it's not clear how it would arise either. For instance, what are white's and black's last moves before such a position? But as you noted, there are some special cases that need to be investigated before I can be certain that a 40 move solution cannot be achieved, and if one is incredibly lucky it could be possible to retrace the game backwards from one such special case position to find a 40 move solution. Are there any other 'safe squares' than b6 and g6 for the black king that need to be investigated? I don't see any.  
  > **username_3**: @username_1 Shaved off another with username_2's idea, now 1 above the lower bound.  
  > **username_3**: @username_2 In your setup you have exactly one column with doubled pawns. As you can see in my solution, we are left with one major piece, which can not be captured without doubling another column (or tripling). That makes me believe my solution is optimal (although it's not a hard proof).  
  > **username_2**: @username_1 *Are there any other 'safe squares' than b6 and g6 for the black king that need to be investigated? I don't see any.* I don't see any either. Only b6 and g6 are safe and only in a setup with a white pawn on c7 (f7) and a black piece on b8 (g8).  
  > **username_2**: @username_3 Good point. So it comes down to the question, whether with 15 pawn captures you can "undouble"/"untriple" all pawns.  
  > **username_2**: @username_3 I would guess you need an even number of captures to do that but have no prroof.  
  > **username_2**: I have a proof now: Start by labeling the columns/ranks by numbers, i.e. a=1, b=2...h=8. Then for each pawn calculate the sum of the respective ranks. So in the initial position this sum is 1+2+3...+8=36. Now, if a pawn captures it changes its rank by "1" (either plus or minus 1). In order to achieve a solution in 40 moves, we have to queen the pawns on each of the squares on the 8th row. This means that in the final position the sum of ranks (for pawns+promoted pawns now) needs to be 36 again. However we also need to do 15 captures with these pawns.  
  > **username_2**: Each capture changing the rank sum by 1, it is impossible to end up with 36 as sum, basically because 15 is an odd number.  (If you have x captures to the left and y captures to the right you end up with equations: x-y=0 and x+y=15 which has no integer solution) That proves that it is impossible to achieve a solution in 40 moves and the solution with 41 moves given by username_3 is optimal.  
  > **username_4**: @username_2 Neat parity argument, but you mean files, not ranks.  
  > **username_3**: @username_2 That's exactly what I meant with my comment, nicely formalized.  
  > **username_5**: I think @username_2 's nice argument should be moved from the comments into the body of the answer, so that it doesn't get lost.  
  
  ---
  
  