A regular game has been played where White still has all 16 pieces. White promoted only a single pawn, and not to queen. In the final position all 16 pieces of White have (at least) one mating move. Yes, also the king. Find the position. ("The"=it is mostly determined, allowing for variations)

Yes, fitting for Silvester there is a little gag hidden, but it is entirely conforming to FIDE rules.


1 Answer 1


This is a solution that comes with a promoted bishop. All 16 white pieces have at least one mating move, the star among them being 0-0-0#.

A problemists problem with this is, as was pointed out by OP, that it's an impossible position due to black's double pawns.

[fen "1nbBbb2/PP1P1PP1/1BpkBQ2/2p1p1N1/1P3P2/2N4R/8/R3K3 w Q - 0 1"]
  • Essentially the same position (as I said, some leeway) as I intended. I merely saved two pawns, but then, I'm a chess composer pro :-) See here: hamburger-schachverband.de/berichte/2022/silvester22.bmp Jan 4 at 9:14
  • I am Patz, you are Gurke, the Pc6 protects against the R mate :P Jan 4 at 10:52
  • Told the webmaster to overwrite it with your correct version :-) Jan 5 at 9:07
  • 1
    <sad horn>As an expert spotted, how did Black get the double pawn?</sad horn> Rats. Jan 8 at 9:25
  • "J'adoube!" was said, I guess. Edited.
    – Pit
    Jan 8 at 13:13

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