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Maximum number of non-pawn pieces on the board notes how to get the maximum number of non-pawn pieces on the board.

But what about the maximum number of points possible? It seems like this starts easily enough. For instance: 1. a4 b6 2. a5 Bb7 3. a6 h5 4. axb7 Nc6 5. b8=Q, with a plan of Ra1-a3-e3 to allow Black's a-pawn to queen.

The basic idea would be to have each side capture the other's 4 minor pieces, so that 1) the capturing pawn becomes passed and 2) it creates an opposing pawn become passed. Once the two pawns above have queened, White or Black could take bxa(1-8), Then cxb(1-8) and so forth.

This is the outline, but how do we create a legal game that satisfies this?

2 Answers 2

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As Jeremy said, the maximum is indeed 182 points. Here is a proof game for it.

[FEN ""]
[startply "110"]

1. b4 g5 2. h4 a5 3. h5 a4 4. h6 a3 5. Bb2 Bg7 6. hxg7 axb2 7. a4 h5 8. a5 h4 9. a6 h3 10. Nf3 Nc6 11. b5 g4 12. b6 g3 13. Nh2 Na7 14. bxa7 gxh2 15. g4 b5 16. g5 b4 17. g6 b3 18. e4 d5 19. c4 f5 20. c5 f4 21. Bc4 Bf5 22. exf5 dxc4 23. Nc3 f3 24. Ne2 fxe2 25. f4 Kd7 26. Kf2 e5 27. f6 e4 28. f5 e3+ 29. Kf3 Kc6 30. f7 Kb5 31. c6 Nf6 32. g8=Q Nd7 33. cxd7 c5 34. f8=Q Qb6 35. g7 Re8 36. a8=Q c3 37. d4 c2 38. d5 c4 39. f6 c3 40. f7 e1=Q 41. d6 e2 42. d8=Q c1=Q 43. Rf1 h1=Q+ 44. Kg4 h2 45. Qb7 Qd5 46. a7 h1=Q 47. a8=Q Qf2 48. d7 b1=Q 49. Qfe7 b2 50. Qh7 Qba2 51. g8=Q b1=Q 52. f8=Q c2 53. Qdc7 Qcb2 54. d8=Q e1=Q 55. Rg1 c1=Q
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I think you can get nine queens and two rooks for each side. Consider this: a4 b5; a5 Bg7; a6 b4; axb7 Nc6; b8(Q) Na5; Na3 bxa3; b4 a2; Rb1 a1(Q).

You have made two captures, promoted two pawns, and cleared the way for two more pawns to advance to the eighth rank. In other words, you can make it so that one pawn capture allows two pawns to queen.

The tools just need to get out of the way.

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