I had an interesting idea recently. While there are many proof games ending in checkmate or stalemate, why not both, I thought. So, thus I wondered, what is the fastest way to reach a position in which one side is forced to give either checkmate or stalemate?

So far, my best result is a bulky 26 moves, or 52 plies. In the end, White must choose between Kb7# and Kc7# or Kc8=.

[FEN ""]

1. a4 b5 2. axb5 g5 3. Rxa7 Ba6 4. bxa6 Bh6 5. f4 Nf6 6. fxg5 Nc6 7. h4 O-O 8. Rxa8 Nd4 9. gxh6 Nf3+ 10. Kf2 Ng5 11. hxg5 Qb8 12. Ke3 Qxb2 13. c4 Qxb1 14. Kd4 Qxc1 15. Kc5 c6 16. Kb6 c5 17. a7 Qxd1 18. d3 Qxf1 19. gxf6 Qxg1 20. g4 Qxh1 21. g5 Qb7+ 22. Kxb7 Rb8+ 23. Kxb8 d5 24. g6 d4 25. g7 e6 26. e4 e5

1 Answer 1


12 moves (24 plies), adapting Loyd's classic 19-ply help-stalemate:

[FEN ""]

1. d3 a5 2. Qd2 Ra6 3. Qxa5 h5 4. Qxc7 Rah6 5. Qxb7 f6 6. Qxb8 d6 7. h4 Bf5 8. Rh3 Kf7 9. Qxd6 Bh7 10. Qe6+ Kg6 11. f4 Qxd3 12. Kf2 Qg3+

and White must choose between 13 Kxg3 stalemate and 13 Rxg3 checkmate.


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