An old self-stalemate puzzle has been popping up in my feed lately. I was unable to do better than the existing answers, so I created a different puzzle from it. For reference: One-sided chess: Can you stalemate yourself?

Goal: Set up all the white pieces so that no piece has a legal move, as close as possible to a reachable position.


  1. Start with both white and black in their normal positions.
  2. End with 16 white pieces on the board.
  3. No piece has a legal move.

The only purpose of the black pieces being on the board is to give white pawns a chance to legally capture. Once you have all the white pawns where you want them, feel free to remove all the black pieces and get to the end game. Or more realistically, find the end position you want, and determine how many pawn captures you need, and never bother with actually having the black pieces on the board.


A perfect score is 0, which is achieved if white has no moves and all 16 pieces are in reachable positions without any captures or promotions. Since captures and promotions are part of a regular game of chess, there is a minimal penalty for each: 1 point per promotion or capture.

The following illegal moves cost more points:

3 points - Major piece out of position:

  • Pawn in 8th row
  • Bishop on wrong color
  • Pawn in location unreachable via capture

5 points - Replace any white piece on the board with another white piece


The lowest score for an answer reachable in a real game is the winner. If no such answer is posted, the winner will be the answer with the lowest score that also contains illegal moves.

  • 1
    $\begingroup$ So you can't use black pieces to block the paths of white pawns, or to prevent the white king from finding a legal square to get to? $\endgroup$
    – msh210
    Jul 27, 2021 at 2:32
  • 1
    $\begingroup$ Correct. The only things that should block white pieces are other white pieces, or the edge of the board. $\endgroup$ Jul 27, 2021 at 5:06

3 Answers 3



2 if I understand the puzzle correctly:
enter image description here

  • $\begingroup$ Wow. I thought my personal answer was pretty good, but this is most certainly better. Not going to be surprised if this is the winner. Technically, the score is positive, not negative, but otherwise, well done. $\endgroup$ Jul 27, 2021 at 5:09
  • $\begingroup$ Nice answer, as always. Good job Albert.Lang =) $\endgroup$
    – justhalf
    Jul 27, 2021 at 5:43

Albert.Lang's answer is optimal.

To prevent knight from moving, you need it in a column with 3 pieces next to it. This requires a column with a single piece. Assuming we want to lock-in one knight without captures, another column needs to contain just a bishop. That bishop needs rooks next to it (anything else escapes through the hole above bishop). The problem is that we cannot use rooks by the knights (one is obvious because it is on row 7, the other requires something on C7/F7 to block knight). Therefore, we need to promote two pawns for new rooks.

Another possible solution: Score 3 - one capture, 2 promotions.


Also, here are the two best solutions I found that don't use black pieces. Both are illegal. This one has score 5 (same color bishops + 2 promotions).


And there is another illegal no-capture option, this one with 4 points (8th rank pawn + promotion):


Neither of these can be improved because you need that many rooks for them.

Finally, it is obvious you can't just replace one knight with another figure - you would need to have 8 pieces in 8th rank and 8 pieces/pawns in 7th + one in 6th = 1 too many. But it is trivial to find solutions with one promotion and score 6.

  • $\begingroup$ The best I had managed prior to posting the puzzle was the 5 point same color bishop. $\endgroup$ Jul 27, 2021 at 13:52

Ok, I'll start... and maybe finish too...

Get white to the reverse of the starting position: all the white pawns on 7th rank, and all the white pieces on 8th rank, in their starting files. This is legal, with zero captures by white pawns. Then replace two knights with something else, say rooks.

Score: -10 (two illegal replacements).

Why there might not be a better score:

I don't think there's a solution where a white knight survives, though I don't know how to prove that yet. If that is correct, then you need illegal moves to get rid of knights and still end up with 16 white pieces. Going through the list of illegal moves, the 'least illegal' is just to replace them at the end for a 10 point penalty.

EDIT: that is not even close to being correct, as Albert.Lang's answer shows :-)

  • 1
    $\begingroup$ I considered putting this in the question as a starting point, as it was, in fact, my first answer. I did better, but totally missed the current best play. $\endgroup$ Jul 27, 2021 at 5:11

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