4
$\begingroup$

Here's simple and humourous chess puzzle that I created after a few non-consecutive hours. I had to prove the concept feasible while making it as economic as possible.

White To Move, And Selfmate In One Move:

board

For Those Who Don't Know A Selfmate Is: Wikipedia

Many thanks to LeppyR64 for suggesting that I add that link in.

A Clarification: The board is from White's POV, so the black pawn e7 is on it's starting square, and thus it does not block the White King from moving to d8 tried in Zymurge's answer.

And remember, every piece has been meticulously placed with a reason. Every piece plays a role, no matter how small or useless it may seem.

(My phone just seems to hate picture uploading to the app. Should I contact the developers? It seems to be a problem.)

There is more to this than meets they eye. The obvious solution of 1. Bxe7 Nc7# or Bxf7# is just a sideline.

Can you solve the mystery of what the solution is?

$\endgroup$
  • 1
    $\begingroup$ Now that I've solved it, I'm wondering why you declare Bxe7 a sideline solution when it opens up two forces for black: Nc7# and BxF7#. Either way, it's a self-mate $\endgroup$ – karnesJ.R Jun 12 at 23:04
  • $\begingroup$ When compared to the theme of what I was going for here, it is a sideline. It's the cost at which the theme was made possible. $\endgroup$ – Rewan Demontay Jun 12 at 23:06
1
$\begingroup$

To submit something that no one has yet suggested, Rb2..Nc7# is valid but not mandatory for black because the option of Bxf7 is always in play unless black's king is placed into check.

The solution hinges on F7 because outside of A1, it is the subject square of which Black has the most options.

A1 tells us two things when combined with F7:

  • You must check the black king
  • You must use white's F7 Pawn to eliminate G8 which always has the freedom to take F7

This makes my official solution as follows:

White's First Move

Pxg8.. Promote to Black Pawn

Why?

This removes the only other piece with freedom (aside from the knight at a1). We are left with a connundrum though. Any other piece we promote illegally has the option of moving (Nf6) or capturing (?xf8) which doesn't force Nc7#. By choosing a black pawn, we paralyze the queen.

Black is Forced to

Nc7#

Why?

Going from left to right on the board: A1 is free to move, but only onto C7. B6 cannot progress due to B5. E5 through E7 is column locked and stuck by E8. G8 (now a pawn) is blocked by G7 which is blocked by G6 which is pinned by E8. H8 and H7 are blocked in. H6 is pinned by H5

Alternatives:

Since you're now illegally promoting pieces you can actually pick any black piece. Rf8#, Qf8#, Qf7#, Bf7X, Nf6# (My favorite as I didn't see it before). In any situation, promoting illegally to a black piece does force a checkmate on the next move as black has no options that don't result in an immediate checkmate.

Very fun puzzle. Required no less than 10 minutes of analysis.

$\endgroup$
  • 1
    $\begingroup$ Thanks, and nice analysis. Although your answer was figured out before (see jafe's answer), I like your in depth analysis. And an Ilegal promotion to a black king works the same as a black pawn. $\endgroup$ – Rewan Demontay Jun 12 at 18:20
  • $\begingroup$ At the time I had started typing there was only Zymurge and Sid's answers. I had to step away from the keyboard and submitted my answer late. On one hand I'm sad to not be first, but on the other, it is good to be right. $\endgroup$ – karnesJ.R Jun 12 at 18:24
  • 1
    $\begingroup$ It's all good. It took me a few hours to create my position and make it work. I wanted it so that all possible six illegal promotions worked with the same pawn. Plus it needed to be a legal position and as economic as possible. $\endgroup$ – Rewan Demontay Jun 12 at 18:31
  • 1
    $\begingroup$ Answer accepted! ; D $\endgroup$ – Rewan Demontay Jun 12 at 18:54
  • 1
    $\begingroup$ I never thought to promote to a second King. Mainly because I thought about breaking the rules, not just throwing them out the window entirely. I did consider the possibility of simply choosing to not promote, but it would still leave Black with any move that satisfied ?xg8 instead of forcing Nc7#. When I realized that only a black piece in that square would force Nc7# I picked the pawn because it wouldn't be able to move at all thus forcing only Nc7#. $\endgroup$ – karnesJ.R Jun 12 at 22:32
4
$\begingroup$

The answer is:

Bxe7 then either Nc7 or Bxf7 is mate

This works because

Black only has two pieces that can move, which are the two mating moves above. The trick is that each move individually cannot be removed without also removing the subsequent mate. If we go Rxa8 to remove the knight, then the king escapes the bishop's check where the rook vacated. If the pawn takes the bishop, black has other non-mating options aside from the the knight move.

The key to moving the white bishop is that it allows a discovery check from the black queen when the black bishop is moved out of row 8.

Clever puzzle in that the solution is not the obvious elimination of options. Kudos for my son for helping me to find the answer :-)

$\endgroup$
  • $\begingroup$ Nice try, but that does not work. If the rook takes the Knight, Bxf7 IS NOT checkmate because the White King can escape to where the rook was. Every piece has been place meticulously for a reason $\endgroup$ – Rewan Demontay Jun 12 at 2:54
  • $\begingroup$ The board is from White's POv, so the black pawn d7 is on it's starting square, and thus it does not block the White King. $\endgroup$ – Rewan Demontay Jun 12 at 3:01
  • $\begingroup$ That was my bad on mixing up the white POV. I caught that when I was explaining my so-called solution to my son! I edited to what I think is now the correct solution, which he basically found with a little of my steering. $\endgroup$ – Zymurge Jun 12 at 3:14
  • $\begingroup$ Whoa, wait a minute ... upon re-reading you straight up said that my answer is not the answer and is just a sideline? How does it not work for a one move self-mate? $\endgroup$ – Zymurge Jun 12 at 3:16
  • 1
    $\begingroup$ I checked it with Popeye. Zymurge's solution of 1 Bxe7 Nc7#/Bxf7# is the unique solution. What did you intend, @RewanDemontay? BTW Black has 3 knights. Is the position in the question what you intended? $\endgroup$ – Rosie F Jun 12 at 9:02
3
$\begingroup$

Since we're encouraged to think outside the box,

fxg8 and illegally promote to any black piece. (Queen, rook, bishop, knight, king and pawn all work.) Black has no legal moves except to give mate, either with the newly promoted piece or with the knight in the corner.

This sort of "solution" is not without precedent. For example,

Wikipedia's Joke chess problem article contains an example where the "solution" is to promote to an opponent's piece.

$\endgroup$
1
$\begingroup$

Alongwith what Zymurge found ,

1. fxg8=N Nc7#
This also mates in 1 although i am not sure if it is fundamentally as different from what was already found.

$\endgroup$
  • $\begingroup$ That does not work as black can capture the newly born knight with their Queen or King. But you do have the right idea. $\endgroup$ – Rewan Demontay Jun 12 at 12:21
  • $\begingroup$ Yes, except that why should you capture if you can checkmate? $\endgroup$ – Sid Jun 12 at 12:24
  • 1
    $\begingroup$ Because it's a SELFMATE. White must force black to checkmate him. If Black has an option to not checkmate White, then they will choose not to checkmate White. $\endgroup$ – Rewan Demontay Jun 12 at 12:26
1
$\begingroup$

I'm fairly certain

Bxg7 is also a valid self-mate

Going through all possibly moves:

Nc7 is checkmate, Nb6 is blocked by the pawn. The Knights on g6 and h6 cannot be moved without putting black in check. All of black's pawns are blocked. Bxf7 is checkmate, and the Queen cannot be moved.

The only piece I'm confused about the purpose of is the white pawn on h4.

EDIT:

The white pawn on h4 exists to prevent the Queen from moving down, which would also be a self-mate.

$\endgroup$
  • $\begingroup$ Unfortunately, that wouldn't work because then the Black king or queen could take the bishop without checkmate. $\endgroup$ – Rewan Demontay Jun 12 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.