Sam Loyd (141-1911) created the shortest (?) chess game ending in stalemate: 19 ply: 1.e3 a5 2.Qh5 Ra6 3.Qxa5 h5 4.Qxc7 Rah6 5.h4 f6 6.Qxd7+ Kf7 7.Qxb7 Qd3 8.Qxb8 Qh7 9.Qxc8 Kg6 10.Qe6 stalemate. A variant of the same length is 1. c4 h5 2. h4 a5 3. Qa4 Ra6 4. Qxa5 Rah6 5. Qxc7 f6 6. Qxd7+ Kf7 7. Qxb7 Qd3 8. Qxb8 Qh7 9. Qxc8 Kg6 10. Qe6. A 24-ply stalemate with NO CAPTURES, also by Loyd: 1. d4 d6 2. Qd2 e5 3. a4 e4 4. Qf4 f5 5. h3 Be7 6. Qh2 Be6 7. Ra3 c5 8. Rg3 Qa5+ 9. Nd2 Bh4 10. f3 Bb3 11. d5 e3 12. c4 f4. As far as I know the minimal length of Loyd's stalemates has not been proven, but seems fairly plausible. I do not know whether a "double stalemate" has ever occurred in a real game. A composed example in 36 ply (Enzo Minerva 2007): 1. c4 d5 2. Qb3 Bh3 3. gxh3 f5 4. Qxb7 Kf7 5. Qxa7 Kg6 6. f3 c5 7. Qxe7 Rxa2 8. Kf2 Rxb2 9. Qxg7+ Kh5 10. Qxg8 Rxb1 11. Rxb1 Kh4 12. Qxh8 h5 13. Qh6 Bxh6 14. Rxb8 Be3+ 15. dxe3 Qxb8 16. Kg2 Qf4 17. exf4 d4 18. Be3 dxe3 and now neither side could legally move. This record might well be beatable by finding one shorter than 36 ply. I posed to some of my friends the question of whether a double stalemate is possible with no captures -- or if captures are necessary, then use as few as possible, and preferably of lower-average mobility men (chessmen ordered by increasing average mobility: PNBRQ) -- and try to create the shortest possible double stalemate game subject to those restrictions. We searched for quite a while, whereupon Monty McGovern found a good looking final position, which I then was able to prove was (a) reachable in a 24-move (48 ply) game, and (b) not reachable in any fewer than 48 ply. This game involves a single capture of a knight by each side, and hence is pretty close to "optimality"... we do not know if a double stalemate is possible using (1,0) or (0,0) captures, or whether you can do it with captures of pawns rather than of knights but I find it plausible those tasks are impossible. Here is the game: Monty McGovern and Warren D. Smith, April 2014 1. b4 Nf6 2. Bb2 h5 3. BxNf6 g5 4. g4 Bg7 5. Be5 b5 6. Bg3 a6 7. Bg2 Bb7 8. Nc3 BxNc3 9. Nf3 Nc6 10. h3 Bd4 11. Nh2 Bb6 12. e4 d4 13. Ke2 Kd2 14. Qg1 Qb8 15. f3 Na7 16. Rf1 Rc8 17. Be1 c6 18. e5 d4 19. Kf2 Bd8 20. a4 Kc7 21. a5 h4 22. c4 f5 23. e6 d3 24. c5 f4 double stalemate. The following URL should show you a picture of the final position: http://www.apronus.com/chess/wbeditor.php?p=A____BRQR___P_KBN___p_P_P_P___pP... The proof that 48 ply is shortest possible to reach this position is roughly as follows. You simply add up the shortest-path move-count distances of each piece from its starting to final square (taking account of never-moving pawns as obstacles), plus observe that the Be1 and Kf2 cannot both travel on shortest paths so 1 extra ply needed. That yields a lower bound of 22+22=44 ply. Then you note that the missing Knights have to die somehow, which requires at least 2 extra ply per Knight, because the Ns have to get onto some enemy's shortest path or that shortest path has to gain one extra hop and the N has to get onto that path... reaching 22+22+2+2=48 ply. You also have to take into account the possibility of castling or of using "the other knight" than you think, or of one of the Ns capturing the other... but it is not hard to see these options cannot decrease us below 48. -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)