3
$\begingroup$

In standard chess rules, a pawn is promoted if it reaches the far side of the board. Usually, the pawn's owner chooses to promote it to a queen.

Is it possible for 18 queens to be on the board at once, in a legal game of chess? (Two original queens, plus all promoted pawns). Assume the players co-operate to allow this. However, if the game reaches checkmate, it must stop.

$\endgroup$
0
$\begingroup$

While definitely not as orderly as my last solution, it seems this works. Not sure if it can be improved in terms of number of moves, there are quite a few non-pawn moves in there (required for the captures and avoiding checks). EDIT: Actually, it seems this is a solution to the exact same problem, which solves it in 96 half-moves (as opposed to my 139).

PGN:

1. e4 d5 2. e5 d4 3. e6 Bd7 4. d3 f5
5. exd7+ Kf7 6. Be3 dxe3 7. d4 e5 8. d5 Qe7
9. d8=Q e2 10. Kd2 e1=Q+ 11. Kd3 e4+ 12. Kd4 e3
13. f4 e2 14. d6 Q1h4 15. d7 e1=Q 16. Qc8 Nf6
17. d8=Q b5 18. Bd3 b4 19. Be4 fxe4 20. c4 Qeg3
21. c5 a5 22. c6 Ra7 23. Nc3 bxc3 24. b4 Rb7
25. cxb7 Na6 26. b8=Q a4 27. b5 Nc5 28. b6 Nb3+
29. axb3 a3 30. b7 a2 31. Qa8 c2 32. b8=Q c1=Q
33. b4 c6 34. b5 e3 35. Rb1 Qca3 36. Rb2 a1=Q
37. b6 e2 38. b7 e1=Q 39. Qba7 g5 40. b8=Q Nd7
41. Ne2 Ke6 42. Ke4 c5 43. Rd2 c4 44. Rd3 c3
45. Rf3 c2 46. Rhf1 c1=Q 47. h3 Qh2 48. g3 g4
49. Qe8 gxf3 50. Rg1 f2 51. g4 f1=Q 52. g5 Q4f2
53. g6 h5 54. g7 Qcb2 55. Qed8 Qec5 56. f5+ Kf7
57. f6 Be7 58. Rh1 Kg6 59. f7 Rh7 60. g8=Q+ Rg7
61. f8=Q Qhg3 62. Rg1 Qgg5 63. Rg4 hxg4 64. h4 g3
65. Qdd5 Ne5 66. Qdb7 Nf7 67. h5+ Kf6 68. h6 g2
69. h7 g1=Q 70. h8=Q

Again, animation welcome.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for putting in the effort again! It does seem to be a duplicate question though, so I have marked it as such. $\endgroup$ – TenMinJoe Dec 4 '17 at 21:09
  • $\begingroup$ No problem - I liked doing it anyway! $\endgroup$ – Lolgast Dec 4 '17 at 21:19

Not the answer you're looking for? Browse other questions tagged or ask your own question.