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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!

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  • $\begingroup$ 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
    Commented Feb 23, 2020 at 3:34
  • $\begingroup$ Black must give the mate immediately after White's underpromotion. $\endgroup$
    – Rewan Demontay
    Commented Feb 23, 2020 at 3:57
  • $\begingroup$ Oh, also did you mean 7th rank instead of 2nd? I overlooked that when I made my edits $\endgroup$
    – HTM
    Commented Feb 23, 2020 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
    Commented Feb 23, 2020 at 14:21
  • $\begingroup$ @Bass That is correct, the solution must be unique. $\endgroup$
    – Rewan Demontay
    Commented Feb 23, 2020 at 14:47

5 Answers 5

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N:9 B:9 R:8 Q:9 P:13

After I'd posted 4 problems, Rewan Demontay revealed that actually work on this had continued in another forum a couple of years ago, and better problems were eventually found, but never posted here.

To avoid more people wasting time like myself & poor @happystar, I have posted the 4 all-time records here, so that people can see. In my opinion, it's good work: none of the four are likely to be improved further.

My only novelty is a new 5th case: "dummy pawn" (1862 rules, pawns can stay unpromoted on the 8th rank). See https://chess.stackexchange.com/questions/21212/when-if-ever-was-it-a-rule-that-pawn-promotion-was-optional for details of this amazing rule. So this is how Queen Victoria would have played chess in the latter half of her reign :) Maybe someone can improve it? There are a number of different matrices possible.

All five White promotions can be forced uniquely (with multiple Black responses allowed - this is normal self-mate protocol). All these are proved sound by Popeye solving software.

Note that all 4 of the previous problems still remain even if dummy pawn is available as an additional promotion target.

N: 9 units (Hauke Reddmann)

enter image description here

B: 9 units (Rewan Demontay)

enter image description here

R: 8 units (Rewan Demontay)

enter image description here

Q: 9 units (Hauke Reddmann)

enter image description here

P: 13 units (Laska, Retudin, Rewan Demontay) non-standard material:RB

enter image description here

P: 14 units (Laska)

enter image description here

The P task is necessarily trickier to realize so is likely to require more units.

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6
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EDIT:

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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  • 1
    $\begingroup$ Could the downvoter please explain their reason for doing so? $\endgroup$
    – Bass
    Commented Feb 23, 2020 at 13:31
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    $\begingroup$ The 10 unit position does not count because the promotion is not unique, which violates the rules. $\endgroup$
    – Rewan Demontay
    Commented Feb 24, 2020 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
    Commented Feb 24, 2020 at 7:30
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Consider the following solution:

14 pieces

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

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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    $\begingroup$ 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
    Commented Feb 23, 2020 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
    Commented Feb 23, 2020 at 17:10
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I can get 8 pieces for promotion to Rook, an improvement of 2 pieces over Laska's solution. I have no improvement for promotion to Bishop Knight or Queen

enter image description here

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    $\begingroup$ Well done. Unfortunately all of the recent ones are anticipated by problems that were done in the last couple of years, but were never posted here. See the links in Rewan’s comment to my post $\endgroup$
    – Laska
    Commented Aug 1, 2023 at 13:15
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    $\begingroup$ I have posted Rewan & Hauke's problems in my own answer. $\endgroup$
    – Laska
    Commented Aug 1, 2023 at 13:36
1
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Dummy pawn promotion (flawed as explained in comment)

allows for a 13 piece solution. (At least I think the 'promotion' to pawn is the only self mate in 5.)

enter image description here

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7
  • $\begingroup$ What are the lines considered here? (apart from the main line of pawn race at b and h) $\endgroup$
    – justhalf
    Commented Aug 1, 2023 at 21:19
  • $\begingroup$ Hi this nearly works, and is very clean. I like bPh1 which is very helpful for the solver. There is a cook b2-b3-b4-b5-b6-b7 however. $\endgroup$
    – Laska
    Commented Aug 2, 2023 at 5:38
  • $\begingroup$ I believe 2N4b/N2kPPn1/N6p/8/8/8/BP6/KB5p w - - 0 1 is a fix. $\endgroup$
    – Rewan Demontay
    Commented Aug 2, 2023 at 7:18
  • $\begingroup$ @RewanDemontay I don't think it quite works: 1. b4 ! threat: 2.e8=Q+ Sxe8#. 3B3b/N2kPPP1/7p/2P5/8/8/BP6/KB5p does work I think, but still 14 units (and non-standard material). Maybe some further economization is possible $\endgroup$
    – Laska
    Commented Aug 2, 2023 at 8:58
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    $\begingroup$ @Laska : nice (glad my efforts led to something) $\endgroup$
    – Retudin
    Commented Aug 6, 2023 at 10:16

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