I am posting the present question since here Very unbalanced Chess Positions I have introduced three open problems (i.e., the third one, the fourth one, and the sixth one from the above), regarding maximum material imbalances reachable in final chess positions, that still remain unsolved.
In details, we assume by hypothesis that
Pawn := (P) = 1 point;
Knight := (N) = 3 points;
Bishop := (B) = 3 points;
Rook := (R) = 5 points;
Queen := (Q) = 9 points;
King := (K) = 0 points.
Then, let me quote that "A stalemate occurs when the player on move is not able to perform any legal move, and at the same time is not threatened by check. Thus, stalemate can be forced by both players" (i.e., we can face only two possible stalemate scenarios, white stalemated or black stalemated).
Now, under the standard FIDE rules, our three goals (unsolved problems by 1) are as follows:
First open problem: Which is the maximum difference between the "total score of White" minus the "total score of Black" that black can force by playing a move that results in white stalemated?
Proven lower bound for First open problem is 81, that is achieved through the following play:
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1. g4 f5 2. gxf5 g5 3. h4 g4 4. h5 Kf7 5. f6 Ke6 6. fxe7 Kf6 7. e8=Q g3 8. Nf3 g2 9. Rh4 g1=B 10. a4 b5 11. axb5 a6 12. bxa6 Bb7 13. axb7 Nc6 14. bxa8=Q Qe7 15. Qeb8 Qd6 16. e4 Kg7 17. e5 Qd5 18. e6 Qe5+ 19. Qe2 Qf6 20. exd7 Qf7 21. d8=Q Nce7 22. Qdxc7 Nf6 23. Qea6 Nfd5 24. Qac8 Nf6 25. Ne5 Ned5 26. f4 Kg8 27. Nc6 Ng4 28. c4 Ne5 29. fxe5 Qf6 30. e6 Qf7 31. e7 Qf6 32. e8=Q Qf7 33. Nd4 Qe7+ 34. Kd1 Qf7 35. Qea4 Qg7 36. Q4a7 Qf7 37. c5 Qe6 38. Qd6 Qe7 39. c6 Qf7 40. c7 Qe7 41. Qcb7 Qf7 42. c8=Q Qe7 43. Qdc7 Qf7 44. Nf5 Nf6 45. d4 Ne4 46. d5 B1c5 47. Ne7+ Kg7 48. Bg5 Qg6 49. Nc6+ Qf7 50. Bd8 Nf6 51. Re4 Nd7 52. Re7 Kg8 53. Rxd7 Kg7 54. Ne7 Bb4 55. Ra6 Ba3 56. Rb6 Bc5 57. Ba6 Ba3 58. b4 Bb2 59. b5 Ba3 60. Nd2 Bb4 61. Nb3 Qe6 62. h6+ Kf7 63. Kc2 Rg8 64. Kd3 Rg7 65. hxg7 Kf6 66. g8=Q Ba3 67. Qg1 h5 68. Qgc5 h4 69. Q5c6 Kf7 70. Kd4 Qh6 71. Nc5 Qf4+ 72. Qxf4+ Kg7 73. Qfd6 h3 74. Qf4 h2 75. Qxh2 Kf7 76. Qhc7 Kg7 77. Ke5 Kf7 78. Kd6 Kf6 (1/2, 1/2).
Second open problem: Which is the maximum total value of the white pieces on the board that can be achieved by playing a regular FIDE chess game that ends with white stalemated?
Proven lower bound for Second open problem is 91, that is achieved through the following play:
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1. g4 f5 2. gxf5 g5 3. a4 b5 4. axb5 a6 5. bxa6 Bb7 6. axb7 g4 7. bxa8=Q g3 8. Nf3 c6 9. h4 g2 10. h5 h6 11. c4 d5 12. cxd5 g1=B 13. dxc6 Kf7 14. c7 Qd7 15. cxb8=Q Qd8 16. d4 Qd7 17. Bxh6 Bg7 18. Bg5 e6 19. Qc2 Rh6 20. Qcc8 Rg6 21. d5 Qe7 22. d6 Qf6 23. d7 Qe5 24. d8=Q Qf6 25. Qda5 Qe5 26. Q5a7+ Qc7 27. Bd8 Rg5 28. h6 Rg6 29. h7 Rg5 30. h8=Q Rg6 31. e4 Rg5 32. Ba6 Bf8 33. Qh2 Qe7 34. Qhc7 Kg7 35. fxe6 Qf7 36. e7 Qf6 37. e8=Q+ Qf7 38. Nfd2 Rg6 39. f4 Rg2 40. e5 Rg6 41. e6 Rg5 42. Qeb5 Rg6 43. Q5b7 Rg4 44. e7 Nf6 45. e8=Q Nd5 46. Rh6 Nb6 47. Rxb6 Rg6 48. Qec6 Qd7 49. f5 Kh6 50. f6 Kg5 51. f7+ Kg4 52. Nc3 Bg7 53. f8=R Rf6 54. Re8 Kg5 55. Rd1 Bc5 56. Nde4+ Kf5 57. Rxd7 Ba3 58. b4 Bf8 59. Kd2 Kg4 60. Kd3 Kf5 61. Kd4 Kg4 62. Kd5 Kf5 63. Ree7 Kg4 64. Nc5 Rf1 65. Kd6 Rd1+ 66. Nd5 Kf5 67. b5 Rd2 (1/2, 1/2).
Third open problem: Which is the maximum value of White pieces plus Black pieces on the board that is achievable by ending the game in (white or black) stalemate (i.e., a generic stalemate ending a playable chess game and nothing else)?
Proven bound for Third open problem is given by the interval [168, 180] (i.e., the maximum value of all the pieces on the board, adding the pieces of both players, belongs to the set {168, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180}), since the following game could be played:
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1. a4 b5 2. axb5 c5 3. d4 c4 4. Bd2 c3 5. d5 Nc6 6. dxc6 Ba6 7. b6 d5 8. b7 f5 9. Bf4 h5 10. Bd6 exd6 11. Nd2 g5 12. b8=Q Rh7 13. f4 cxd2+ 14. Kf2 Rd7 15. c7 d4 16. e4 d3 17. Qe2 d5 18. c8=Q d4 19. b4 Kf7 20. b5 Kg6 21. b6 Nf6 22. b7 Nd5 23. Qbc7 Nb4 24. b8=Q d1=Q 25. e5 d2 26. e6 d3 27. e7 Qb1 28. e8=Q+ Kf6 29. c4 d1=Q 30. c5 d2 31. c6 Qdb3 32. h3 Bb7 33. Qa5 Rd5 34. c7 Qe7 35. Qcd8 Re5 36. c8=Q Bd5 37. Qdb6+ Re6 38. Q2b5 d1=Q 39. Qb8b7 Nc6 40. fxg5+ Ke5 41. g6 f4 42. g4 Qd6 43. g5 Qba3 44. Be2 Qdb3 45. Bg4 hxg4 46. g7 f3 47. g8=Q Q1a2+ 48. Kg3 f2+ 49. Kh2 Qdb4 50. Ne2 f1=Q 51. Rc1 Nd4 52. h4 Rb8 53. Rc7 Qfa1 54. h5 g3+ 55. Kh3 g2+ 56. Kh4 g1=Q 57. h6 Qgb1 58. h7 Qb3b2 59. g6 Ke4 60. h8=Q Nb3 61. Qbd7 Bh6 62. g7 Bc1 63. Qgf8 Nd2 64. g8=Q Q4b3 65. Qac5 Kd3 66. Ng3 Re4+ 67. Qdg4 Rc4 68. Qed8 Rc2 69. Qde8 Bc4 70. Qhh7+ Kc3 71. Qbe6 a5 72. Qba6 a4 73. Qaa5+ Rb4 74. Qe1 (1/2, 1/2).
I have found the given bounds by performing "manual" calculations only, but it would be fantastic to have a program to do this kind of stuff by evaluating millions of positions.
