I came up with this rather amusing and bizarre stalemate where Black stalemates itself by self-smothering.

  [FEN "8/8/8/8/8/3p4/pp6/krb1K3 b - - 0 1"]

  1...d2+?? 2. Kd1

The peculiar feature of this stalemate is that Black is unable to move not because moving a piece would put its king into check, but moving a piece would require it to capture its own piece, which is disallowed. This is an example of a single self-smothered stalemate.

The question is, is it possible to have a double self-smothered mate, such that both sides do not have any moves at all because the moves would require capturing of the side's own pieces?

One possibility suggested by the user supercat is the following -

      [FEN "KN1b4/RPpPp3/P1P5/4P3/8/3p1p1p/3PpPpr/4B1nk w - - 0 1"]

      1. e6

And both sides are stalemated. However, this position cannot occur in practice, because you cannot have a Black bishop trapped on d8 with the c7 or e7 pawns still on their original squares.


5 Answers 5


I think I found a double-self-smothered-stalemate that could occur via sequence of legal moves from the normal starting position.

 [FEN "5bnk/4ppbp/4P2p/5P1P/p1p5/P2p4/PBPP4/KNB5 w - - 0 1"]

  1. c3 f6

Each player has seven pawns and two dark-square bishops, so one pawn from each player must have been promoted. Once White's b pawn has captured to a3 and Black's g pawn has captured to h6, White's g pawn and Black's b pawn can promote without any other pawn having to leave its file. I haven't worked out the exact "game", but I see no reason to believe it wouldn't be relatively trivial, particularly if one isn't worried about achieving the position with a minimal number of moves.


Here's a simpler mutual smothered stalemate (9 men on a side, no promoted pieces):

[Title "Mutual smothered stalemate"]
[fen "5brk/4p1pb/4P1p1/6P1/1p6/1P1p4/BP1P4/KRB5 w - - 0 1"] 

Reduce number of units from Noam's record of 18 to a new record of 15.
(8+7 units, no promoted units necessarily on the board, pawn captures all achievable):

[Title "Minimal mutual smothered stalemate"]
[fen "5brk/4p1pb/4P1p1/6P1/6PB/6PK/6PP/8 w - - 0 1"]

Is this the best possible?

It's interesting that this solution is asymmetric. If you wrap both kings with pawns on opposite flanks, it requires 16 units.

EDIT: Inspired by a comment, I want to show more deeply that the position is legal. And if fact prove that, despite appearances at least one promotion took place in the history of the game. Assume first that no promotions happened.

White c-h pawns could have reached their files with 8 captures, which is OK since Black has 7 units left.

What about the Black pawns? All but 1 Black unit were captured by White c-h pawns. BfP ("Black f pawn") captured to g6, as this pawn couldn't have come from h7. Most efficient if BhP captured to g file, to be captured in its turn.

White pawns made 1,2,2,3 captures on files d,e,f,g respectively. So Black b,c,d pawns would have to make a total of 5 captures to get far enough (to d & e files). Together with the 2 captures onto the g file, Black made at least 7 pawn captures. This is making the most efficient assumption that BaP was the one missing Black unit not captured by a WP.

Black made 7 pawn captures, and White has 8 units remaining, so at first it looks OK. But none of these captures were on files a or b, where WaP & WbP began. One of those two WPs must have also got hit by a BP, but how? The only capture possible is of BaP.

Therefore, we have proved our original assumption wrong: at least one promotion happened in the course of the game. That might have been White or Black. E.g. WbP promotes on b1, or e.g. BbP captures only once, and promotes on c1.

As per @Laska"s given permission in the comments, here is a 35-move (optimal?) proof game created by @Rewan Demontay, of their position. The intention of it is that it was made to prove their retrograde analysis of the game history of their position.

[FEN ""]

1. Nc3 h5 2. Ne4 h4 3. Ng3 hxg3 4. fxg3 Nc6 5. Kf2 a5 6. Kf3 a4 7. Kg4 Nh6+ 8. Kh3 Ne5 9. b3 Nf3 10. exf3 Rg8 11. Bd3 Ng4 12. fxg4 d5 13. Be4 dxe4 14. Qf3 e3 15. dxe3 Kd7 16. Qxb7 Ke6 17. Ne2 Kf6 18. Bb2+ Kg6 19. Bf6 Kh7 20. Bh4 Kh8 21. Nf4 Bf5 22. Rhf1 Bh7 23. Ng6+ fxg6 24. Rad1 c6 25. Rd5 cxd5 26. c4 axb3 27. cxd5 bxa2 28. Rf4 Qd6 29. Qb5 Qe6 30. dxe6 a1=Q 31. Qc5 Qa4 32. Qd5 Qxf4 33. exf4 Ra5 34. Qg5 Rxg5 35. fxg5

As you can see, a promotion is needed as @Laska rightfully claimed.

EDIT: Thanks @Rewan. And here is a candidate maximal mutually smothered checkmate:

[Title "Maximal mutual smothered stalemate"]
[fen "4brkq/3p1brb/3Pp1p1/4P1P1/1p1p4/1P1Pp3/BRB1P3/QKRB4 w - - 0 1"]

Legal position with 26 pieces: 6 minor pieces captured & 4 promoted bishops. Some flexibility in the diagram: in keeping with the maximality I've chosen major pieces rather than knights in the back row. Also, kept the smothered kings out of the corners.

Note the pawn moves d2-d3 and e7-e6 unlock the two cages.

Here is a proof game for the 26 unit position.

[FEN ""]

1. h4 a5 2. h5 a4 3. h6 a3 4. b3 g6 5. Bb2 Bg7 6. hxg7 axb2 7. a4 h5 8. a5 h4 9. Rh3 Ra6 10. c4 f5 11. Nc3 Nf6 12. Ne4 Nd5 13. cxd5 fxe4 14. Rc3 Rf6 15. f4 c6 16. a6 h3 17. a7 h2 18. a8=B h1=B 19. g8=B b1=B 20. d6 e3 21. Bc4 Bf5 22. Bd3 Be6 23. Qc2 Qa5 24. Qb2 Qe5 25. O-O-O Bg8 26. Bb1 Bh7 27. Ba2 O-O 28. Rc2 R6f7 29. Qa1 Qh8 30. Rb2 Rg7 31. Kb1 c5 32. Rc1 b5 33. g4 b4 34. g5 Bcb7 35. Bg2 Bd5 36. Be4 Bf7 37. Bc2 Be8 38. Bd1 Bd5 39. Nf3 Bdf7 40. Be4 Nc6 41. Bec2 Ne5 42. Nd4 cxd4 43. fxe5 e6 44. d3

In response to an your original question about whether anything interesting has been written about stalemates, it may be worth mentioning that puzzle creator Sam Lloyd is often credited with two interesting stalemate "games" (though see game 3679 at http://www.chesshistory.com/winter/winter08.html for information about that attribution); in the first, White stalemates Black on the tenth move, after having captured six black pieces:

[FEN ""]

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 

In the second, White is stalemated after Black's twelfth move, while all 32 chessmen are on the board:

[FEN ""]

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

In both of Sam Lloyd's positions, one side is stalemated while the other has many moves available. For both sides to be stalemated would probably take significantly more moves, and for both sides to be smother-stalemated would take more yet.


Just for fun, here is an attempt at optimizing the number of pieces that can be involved. I was able to manage 24 pieces.

[Title "me, chess.stackexchange.com 6/23/2019, mate in 20"]
[FEN "5brk/4p1pb/4p1p1/p3P1Pp/Pp1p3P/1P1P4/BP1P4/KRB5 w - - 0 1"]

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