It’s been awhile since the last one, so cheers for round four!

Given that:

  • It’s White to selfmate themselves in exactly 5 moves, i.e. after White’s fifth move, Black’s only legal move(s) will checkmate White.
  • White has a pawn on the second rank.

Construct a position such that:

  • White’s final move is to promote the second rank pawn to one of a knight, bishop, or rook, while forcing Black to give the mate.
  • The position uses the least number of pieces possible, and is legal.
  • The solution must be unique

The first to post a proper position will get the checkmark. Good luck and good fun to all solvers!

  • $\begingroup$ Just to clarify, does the checkmate have to come immediately after the under-promotion, or can it occur on any of the moves? $\endgroup$ – HTM Feb 23 at 3:34
  • $\begingroup$ Black must give the mate immediately after White's underpromotion. $\endgroup$ – Rewan Demontay Feb 23 at 3:57
  • $\begingroup$ Oh, also did you mean 7th rank instead of 2nd? I overlooked that when I made my edits $\endgroup$ – HTM Feb 23 at 7:30
  • $\begingroup$ Just to clarify: the usual self-mate problem rules still apply, right? In particular, black will resist giving the mate, and the underpromotion is supposed to be the only way to self-mate? $\endgroup$ – Bass Feb 23 at 14:21
  • $\begingroup$ @Bass That is correct, the solution must be unique. $\endgroup$ – Rewan Demontay Feb 23 at 14:47


Managed to pare my score down to 10 pieces (incidentally also worth 10 points in the usual piece value reckoning) by entirely locking down the position, thus removing any last remnants of fun the solver might have found in the earlier versions:

enter image description here

If white starts with 1. c3, black can thwart white's suicidal plans by promoting at b1, which causes an instant stalemate. White's only other legal move is 1. c4, and after that, there are no choices to make, so everything runs on railroad tracks until move 5, when white must underpromote to either a knight or a bishop in order to avoid checkmating black.

Earlier answer versions below.

Here's my attempt at constructing a position where underpromoting the 2nd rank pawn is the only way to self-mate in 5:

enter image description here

As you can see, I've split the problem into 3 parts: a prison for the white king, a prison for the black king, and a prison for the black rook.

The only way to force the black knight to move is to stop the black king from moving around in his prison.

The only way to do that is to promote the h-pawn to a rook and not a queen, as if white chooses a queen then black has ”5. Na3/Nd2” to avoid the checkmate.

Here's the same idea with 12 pieces only, this time requiring a bishop promotion.

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    $\begingroup$ Could the downvoter please explain their reason for doing so? $\endgroup$ – Bass Feb 23 at 13:31
  • $\begingroup$ The 10 unit position does not count because the promotion is not unique, which violates the rules. $\endgroup$ – Rewan Demontay Feb 24 at 3:54
  • $\begingroup$ @RewanDemontay Well, it's your question, so you get to make the rules, or course. Let's just say that I don't really agree with your judgement on disallowing branching, when every possible branch still fulfils every requirement. $\endgroup$ – Bass Feb 24 at 7:30

Consider the following solution:

14 pieces

8/1p6/PP6/8/8/3B4/PPPR1P2/KNRnk3 w - - 0 1

The moves proceed as follows (with all moves forced for black):

1. f4 bxa6
2. f5 a5
3. f6 a4
4. f7 a3
5. f8=<N|B|R> axb2#

The notation in the last line just indicates that white can promote to either a knight, bishop, or rook.

The reasoning is as follows:

Black's only option is to keep advancing their pawn down the board through a series of pushes and captures. White's sixth-rank pawns obstruct Black's pawn in such a way that it can only make one movement down the board. Observe that Black's knight protects the pawn when it finally captures on a2. Black cannot capture a2 with the knight itself, because the knight is pinned to Black's king by White's rook. Black's king is obstructed from moving by White's rook and bishop.

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  • 1
    $\begingroup$ This position has dozens of alternate solutions; pretty much any 5 moves by white will work. I believe OP intended the underpromotion to be the only way to selfmate. $\endgroup$ – Bass Feb 23 at 10:30
  • $\begingroup$ Ah, I see. Since this is my first time doing this sort of puzzle, I based my solution on the definition of selfmate linked by the OP, which doesn't mention anything about a unique solution. Also, in my solution, Black can only mate White in exactly five moves, as opposed to earlier (which is what I thought the OP requested). $\endgroup$ – Synchronous Feb 23 at 17:10

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