This is a kind of a repost of a fascinating chess problem that appeared yesterday. The original poster got shy or something, and bafflingly deleted his own post, which was beginning to attract some good reputation and interest.

In the mean time, I have found some information which I want to share. So I will pose the question and then give one answer. Feel free to improve!

So: background. It seems that someone showed the original poster a problem a year ago, but the poster couldn't solve it, and wasn't quite sure if there was a solution at all. The problem is a selfmate: White to move has to coerce an unwilling Black into checkmating him. The poster didn't know how many moves it would take. Normally selfmates do have a specific move limit: they aren't like studies.

[Title "White to move and selfmate"]
[fen "kbQ5/pb6/P7/3B4/8/7p/7p/7K w - - 0 0"]

Is there a solution? If there is, what is it? And if there isn't, is the problem fixable?

  • be4 then blacks only move is bxb mate
    – yukfoo
    Sep 26, 2020 at 19:34
  • 1
    @youkfoo 1. Be4 Bd5 (Bc6) is allowed. Sep 26, 2020 at 21:57

5 Answers 5


First the verdict on the posted problem. It has certainly no solution in up to 12 moves (checked by Popeye v4.79). And the author of the Gustav software which specializes in selfmates says that there is probably no regular solution to this problem.

Another expert thinks that if White simply captures everything which moves, he can then push bK over to g3 and force Black to win by h3xg2#. That's like taking a long chess problem and solving it like an regular chess game with no limit to the number of moves. Yes it will work to win but it's not really in the spirit of things. That's not to diminish his ingenious approach, but it's just not what we would consider a solution.

The good news is that this problem turns out to be a recognizable example of the Broecker schema, invented in 1891 by Gustav Von Broecker. The German PDB database contains 36(!) other problems which are all based on the same diagonal opposition of bishops. One early very clear example is the following:

    [Title "Broecker - London Chess Fortnightly 1892 - s#9"] 
    [fen "kb4R1/1b6/P7/8/8/7p/6BP/6BK w - - 0 0"] 

    1. Rf8 Bc6 2. Re8 Bd5 3. Rd8 Be4 4. Rc8 Bf3 5. Rh8 Be4 6. Bf3 Bd5 7. Be4 Bc6 8. Bd5 Bb7 9. Bc6 Bxc6#

"s#9" is shorthand for "selfmate in 9" - i.e. White moves first and forces an uncooperative Black to checkmate White by Black's 9th move at the latest.

As the original poster had deduced, there is a key relationship between the square of the queen (or rook) on the 8th rank, with the number of squares separating the bishops. Either player must always make them match if they can, and if he fails to, the other player gets a chance to make them match instead.

White wins because he is able to match by moving 1. Tf8 only. Black has many possible lines of play, but the longest he can hold out is 9 moves. One example solution is shown here.

In the original problem, the position is already a "match" with White to play. He is forced to break the match, and Black can always recreate it, so need never lose. It doesn't seem there is any way for White to use the additional power of the queen to change the parity.

However the matrix has proved very rich over the years, as you can see from the database, and I am sure there's more that can still be mined. The longest problem known to be sound is the following:

    [Title "Rotenburg & Trillon - Die Schwalbe 1980 - s#14"] 
    [fen "k7/1b4R1/N7/7p/8/7p/6BP/6BK w - - 0 0"] 

    1. Rf7! Bc6 2. Re7 Bd5 3. Rd7 Be4 4. Rc7 Bf3 5. Rh7 Be4 6. Bf3 Bd5 7. Be4 Bc6 8. Bd5 Bb7 9. Bc6 h4! 10. Bd5 Bc6 11. Rc7 Bb7 12. Rc1 Bc6 13. Rb1 Bb7 14. Bc6 Bxc6#

Again, here is a sample solution as Black has multiple choices, although White's move is unique at every point.

What's interesting about this one is that Black has a waiting move. This flips the parity, and so normally Black would be able to continue making matches, while White is breaking matches. However, White can flip the parity, by moving to the other side of the main diagonal. This can only be done while ensuring that b7 is never free for bK to enter. Either it must be occupied by bB, or attacked by wR.

  • 2
    "Another expert thinks that if White simply captures everything which moves, he can then push bK over to g3 and force Black to win by h3xg2#." - How would you even go about doing that? Capturing the b8 bishop seems impossible to do without just losing the queen, and capturing the light-squared bishop would be checkmate.
    – D M
    Apr 26, 2018 at 18:53
  • Thanks for your response: here’s one idea. Start with 1. Dh8 Lc6 2. Dh6 & e.g. Ld6! 3. Lxc6+ Kb8 4. Dxd6+ Kc8 5. Dd7+ Kb8 6. Df7 Kc8 7. Dxa7, promote a rook on a8, or 2. ... e.g. Lb7! 3. Dc1 Lc6 4. Da1! Lb7 5. axb7+ and continue with the plan explained above.
    – Laska
    Apr 26, 2018 at 20:26
  • 3
    In your second line, axb7 is checkmate?
    – D M
    Apr 26, 2018 at 21:19
  • 1
    Try instead 1. Qh8 Bc6 2. Qa1 Bb7 3. Bc6 Be5 (best?) 4. axb7+ Kb8 5. Q~ (not 5. Qxe4#?) then force bKc7, promote wP=R, capture remaining bB push bK to g3, and then Rg2+ forcing hxg2#.
    – Laska
    Apr 27, 2018 at 4:04

Is there a solution?

I don't think so. If the White queen goes to a square where it can interpose, then Black can just play BxB.

If, for example, White tries Qe8-d7 or Qe8-e6, hoping Black will move his b8 bishop and allow Qc8+, Black can simply not move the b8 bishop. He could instead just move his light-squared bishop next to White's (White's queen cannot go from d7 or e6 to h8 to take advantage.)

[FEN "kbQ5/pb6/P7/3B4/8/7p/7p/7K w - - 0 1"]

1. Bf3 Bd5 2. Qe8 Bb7 3. Qd7 Be4 (3... Bf4? 4. Qc8+! Bb8 5. Bd5 Bc6 6. Qh8 Bb7 7. Bc6 Bxc6#) 4. Qc8 Bd5

If White does not do that, then I think Black can just use the following plan:

If Qg8, then Black moves bishops 0 apart (that is, BxB.).
If Qh8, then Black moves bishops 1 apart.
If Qc8, then Black moves bishops 2 apart.
If Qd8, then Black moves bishops 3 apart.
If Qe8, then Black moves bishops 4 apart.
If Qf8, then Black moves bishops 5 apart.
If at any time it is not possible to move the bishops like that, then BxB at that moment should force the queen to interpose.

And if there isn't, is the problem fixable?

The problem has a solution if White's queen or bishop start on another square. For example, starting the queen on d7 or g8 would provide lots of places for the queen to go, but only one is correct.

Another possibility is to add a white pawn on a4; moving the pawn loses a tempo and allows White to win. (Although that's not a satisfying way to construct the problem, since it has more than one solution; White could play either a4 or Qh8 as the first move.)

  • I can't seem to understand the problem, could you post an example line shwing the ideas of Qe8-d7 or Qe8-e6? Thanks Nov 10, 2021 at 13:05
  • @DanielAlfredoSottile The queen cannot move off of the eighth rank as then the Black bishop on is free to roam. Nov 11, 2021 at 13:42
  • 1
    @DanielAlfredoSottile I added a line with Qd7.
    – D M
    Nov 11, 2021 at 22:03
  • @RewanDemontay Well, not quite! As I show, if the bishop tries to roam, that can lose. Perhaps this is the key to the actual solution somehow...
    – D M
    Nov 11, 2021 at 22:04

[Edit: D M found an error in my solution. The key reductions are correct but I missed a branch for Black (9. .. Be4) that evades all of those reductions.]

I think I found the solution (16 moves on each side):

[Title "White Self-mate"]
[FEN "kbQ5/pb6/P7/3B4/8/7p/7p/7K w - - 0 1"]

1. Be4 Bc6 ( 1... Bd5? 2. Qh8 Bc6 3. Bd5 Bb7 4. Bc6 Bxc6#) 2. Bf3 Bd5 3. Bg2 Be4 4. Qd8 Bd5 (4... Bc6? 5. Bf3 Bb7 (5... Bd5 6. Qc8 Bc6 7. Be4 Bb7 8. Bd5) 6. Be4 Bc6 7. Qc8 Bb7 8. Bd5 Bc6 9. Qh8 Bb7 10. Bc6 Bxc6#) 5. Be4 Bb7 (5... Bc6 6. Qc8) 6. Bf3 Bc6 7. Bg2 Bd5 8. Qe8 Bc6 (8... Bb7 9. Bf3 Bc6 10. Qd8) 9. Qc8 Bd5 10. Qd8 Bc6 11. Bf3 Bb7 12. Be4 Bc6 13. Qc8 Bb7 14. Bd5 Bc6 15. Qh8 Bb7 16. Bc6 Bxc6#

We look at the pair (q,b) where q is the position of the queen in the 6 possible squares at the top row (1 to 6) and d is the distance between the bishops (1 to 5), after White's turn. The key reductions are:

  1. If (q,b) = (6,1) then done.
  2. If (q,b) = (1,2) then done.
  3. If (q,b) = (2,3) then done.

These are (partially) shown in the first few two branches off the main line. Each reduction in the list can be proven based on the correctness of the preceding reductions. Based on these reductions, we can derive the entire solution. (From move 6 onwards I omit branches admitting a 1-move response that falls under these reductions.)

  • "The key is to look at the pair (q,b) where q is the position of the queen in the 6 possible squares at the top row (1 to 6) and d is the distance between the bishops (1 to 5), when it is White's turn." - Did you mean Black's turn? Otherwise the initial position would satisfy condition 2, and I doubt you meant that.
    – D M
    Nov 11, 2021 at 0:31
  • The positions after Black's move 4 and Black's move 7 are identical... that's a good indication that something went wrong.
    – D M
    Nov 11, 2021 at 0:39
  • No, this doesn't work. 9...Be4 would put the position back to the same as after Black's move 3. See my updated post with Black's plan.
    – D M
    Nov 11, 2021 at 0:51
  • @DM: You are absolutely right; I somehow missed that branch. I'll put a big disclaimer at the top, but if you think I should delete my answer then I will! Thanks! =)
    – user21820
    Nov 11, 2021 at 6:47

I may be missing something, but it seems that white needs their bishop to be on c6 and their queen to be on h8. If one of these is accomplished by changing the setup of the problem, then the other can be accomplished through white's move.


As for fixable ... that rather depends on what the presumed composer wanted to do. Or what history this particular problem had and what possible mistranscriptions it may have passed through during transmission. In extreme cases, it might be a 'place a white piece on the board to make this a self-mate in 5' type of problem that isn't correctly remembered. Or perhaps even a problem solving contest problem that is deliberately incorrect, but with many tries (i.e. moves that would be solutions if it wasn't for just one single move by black that upsets white's idea) that may fool competing solvers.

There are several tries (in 3, 4 or 5 moves) that involve the Bishop on d5. Moving it to e4 makes it a self-mate in 3, at least as far as having a solution is concerned. But it looks as that makes the pawn on h3 without a job, so it is probably necessary to figure out why the composer wanted to have that.

But, as D M already has indicated, replacing the white Queen also does the job, although I would consider it more likely that the position of the Queen is correctly remembered (assuming a mistranscription) -- but that pawn on h3 still seems bothersome ... ?

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