Answers.Body: <p>As per Hauke's invitation, here is an optimal proof game for such a position. This is optimal because a queen must move to give the Black king a square and forty pawns moves must take place. White's h pawn <em>does</em> give checkmate in the end.</p>
  <pre><code>[FEN &quot;&quot;]
  
  1. a4 b5 2. axb5 a5 3. b4 Na6 4. bxa5 Rb8 5. c4 Rb6 6. axb6 c5 7. bxa6 d5 8. d4 e5 9. e4 f5 10. f4 g5 11. g4 h5 12. h4 Qc7 13. cxd5 Qc6 14. dxc5 Bd6 15. dxc6 Be6 16. cxd6 Bd5 17. a7 Kf7 18. fxe5 Ke6 19. a8=Q Bc4 20. b7 Bd3 21. exf5+ Kd5 22. c7 Ke4 23. hxg5 Ba6 24. gxh5 Ne7 25. f6 Kf5 26. e6 Ng6 27. hxg6 Bc4 28. e7 Be6 29. e8=Q Bd7 30. g7 Be6 31. g8=Q Rh6 32. d7 Rh7 33. g6 Rh6 34. g7 Rh8 35. d8=Q Bd7 36. Qexd7+ Kg6 37. f7 Rh6 38. f8=Q Rh4 39. c8=Q Rh7 40. b8=Q Rh8 41. gxh8=Q#
  </code></pre>
  <p>Another proof is that, even with the Black king on b6 or g6, it is impossible to capture all Black pieces in time with only pawn moves.</p>
  <pre><code>[FEN &quot;&quot;]
  
  1. e4 e5 2. f4 f5 3. exf5 Qf6 4. fxe5 d5 5. exf6 Be6 6. a4 Kf7 7. b4 c5 8. fxe6+ Kg6 9. b5 Nd7 10. a5 Ne5 11. a6 Re8 12. b6 Nf7 13. axb7 Bd6 14. bxa7 Re7 15. fxe7 Bf4 16. exf7 Bg5 17. c4 Nf6 18. d4 Rd8 19. cxd5 Kf5 20. dxc5 Rd6 21. h4 Ng8 22. hxg5 Rf6 23. gxf6 h5 24. g4+ Kg6 25. gxh5+ Kf5 26. h6 Kg6 27. h7 Nh6 28. h8=Q Ng8 29. f8=Q Nh6 30. f7 Ng8 31. e8=Q Nh6 32. c6 Ng8 33. a8=Q Nh6 34. c7 Ng8 35. b8=Q Nh6 36. c8=Q Ng8 37. d6 Kf6 38. d7 Kf5 39. d8=Q+ Kg6 40. fxg8=Q#
  </code></pre>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-07-01T08:13:44.267Z
Answers.Text: It took me a while to understand the statement that "it is impossible to capture all Black pieces in time with only pawn moves". It seems to me that you can make all the captures "in time" - what is the obstruction to that? However if you make 15 captures with pawns, then (because 15 is odd) it's impossible to have one pawn on each file at the end of it. So indeed you'll need to waste one further move to have everything promote.
Answers.UserDisplayName: James Martin
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-06-29T16:42:19.500Z
Answers.LastActivityDate: 2021-11-05T18:46:19.910Z
Answers.LastEditDate: 2021-11-05T18:46:19.910Z
Answers.UserDisplayName: Rewan Demontay
Answers.Body: <pre><code>[FEN &quot;3k4/6QP/8/4Q3/8/2Q5/QQ4QQ/RNBKQBNR w KQkq - 0 1&quot;]
  
  1. h8=Q#
  </code></pre>
  <p>This is trivial, as already five queens can threaten the whole board. The only thing you must pay attention to is giving Black a last move. Here it is Kd7-d8.</p>
  <p>Bonus question for Rewan: An answer with the shortest proof game :-)</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2021-06-29T09:25:17.947Z
Answers.LastActivityDate: 2021-06-29T15:00:00.170Z
Answers.LastEditDate: 2021-06-29T15:00:00.170Z
Answers.UserDisplayName: Hauke Reddmann
Body: <p>Is it possible to compose a game where these three conditions are simultaneously satisfied?</p>
  <ol>
  <li><p>No White piece is gone but every Black piece except the king is.</p>
  </li>
  <li><p>All the white pawns are promoted to a queen.</p>
  </li>
  <li><p>On the last move, every square on the board is covered by the White pieces.</p>
  </li>
  </ol>
  <p>Bonus: The last promotion (preferably the h pawn) delivers checkmate.</p>
  
Tags: <problems><checkmate><records><pieces><point-value>
Title: Can there be a checkmate with maximum points of material (103) while covering 64 squares?
ThreadText: # Can there be a checkmate with maximum points of material (103) while covering 64 squares?
  
  (Asked by: username_0 on 2021-06-29)
  
  Is it possible to compose a game where these three conditions are simultaneously satisfied?
  
  1. No White piece is gone but every Black piece except the king is.
  2. All the white pawns are promoted to a queen.
  3. On the last move, every square on the board is covered by the White pieces.
  
  Bonus: The last promotion (preferably the h pawn) delivers checkmate.
  
  **Tags:** problems, checkmate, records, pieces, point-value
  
  
  ---
  
  ## 2 answers
  
  ### Answer by username_1 on 2021-06-29. Score: 14
  
  As per username_3's invitation, here is an optimal proof game for such a position. This is optimal because a queen must move to give the Black king a square and forty pawns moves must take place. White's h pawn *does* give checkmate in the end.
  
  ```
  [FEN ""]
  
  1. a4 b5 2. axb5 a5 3. b4 Na6 4. bxa5 Rb8 5. c4 Rb6 6. axb6 c5 7. bxa6 d5 8. d4 e5 9. e4 f5 10. f4 g5 11. g4 h5 12. h4 Qc7 13. cxd5 Qc6 14. dxc5 Bd6 15. dxc6 Be6 16. cxd6 Bd5 17. a7 Kf7 18. fxe5 Ke6 19. a8=Q Bc4 20. b7 Bd3 21. exf5+ Kd5 22. c7 Ke4 23. hxg5 Ba6 24. gxh5 Ne7 25. f6 Kf5 26. e6 Ng6 27. hxg6 Bc4 28. e7 Be6 29. e8=Q Bd7 30. g7 Be6 31. g8=Q Rh6 32. d7 Rh7 33. g6 Rh6 34. g7 Rh8 35. d8=Q Bd7 36. Qexd7+ Kg6 37. f7 Rh6 38. f8=Q Rh4 39. c8=Q Rh7 40. b8=Q Rh8 41. gxh8=Q#
  
  ```
  
  Another proof is that, even with the Black king on b6 or g6, it is impossible to capture all Black pieces in time with only pawn moves.
  
  ```
  [FEN ""]
  
  1. e4 e5 2. f4 f5 3. exf5 Qf6 4. fxe5 d5 5. exf6 Be6 6. a4 Kf7 7. b4 c5 8. fxe6+ Kg6 9. b5 Nd7 10. a5 Ne5 11. a6 Re8 12. b6 Nf7 13. axb7 Bd6 14. bxa7 Re7 15. fxe7 Bf4 16. exf7 Bg5 17. c4 Nf6 18. d4 Rd8 19. cxd5 Kf5 20. dxc5 Rd6 21. h4 Ng8 22. hxg5 Rf6 23. gxf6 h5 24. g4+ Kg6 25. gxh5+ Kf5 26. h6 Kg6 27. h7 Nh6 28. h8=Q Ng8 29. f8=Q Nh6 30. f7 Ng8 31. e8=Q Nh6 32. c6 Ng8 33. a8=Q Nh6 34. c7 Ng8 35. b8=Q Nh6 36. c8=Q Ng8 37. d6 Kf6 38. d7 Kf5 39. d8=Q+ Kg6 40. fxg8=Q#
  
  ```
  
  > **Comments:**
  > **username_2**: It took me a while to understand the statement that "it is impossible to capture all Black pieces in time with only pawn moves". It seems to me that you can make all the captures "in time" - what is the obstruction to that? However if you make 15 captures with pawns, then (because 15 is odd) it's impossible to have one pawn on each file at the end of it. So indeed you'll need to waste one further move to have everything promote.  
  
  ---
  
  ### Answer by username_3 on 2021-06-29. Score: 11
  
  ```
  [FEN "3k4/6QP/8/4Q3/8/2Q5/QQ4QQ/RNBKQBNR w KQkq - 0 1"]
  
  1. h8=Q#
  
  ```
  
  This is trivial, as already five queens can threaten the whole board. The only thing you must pay attention to is giving Black a last move. Here it is Kd7-d8.
  
  Bonus question for username_1: An answer with the shortest proof game :-)
  
  
  ---
  
  