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As an extension to this question about can a rook and knight without a king stalemate a opposing king, and the traditional minimum-mating-material question, I had a puzzle question:

What are the minimum combinations of material (without a king) required to forcibly mate an opposing king?

  • R + R/Q is clearly sufficient

  • R + N + N/B maybe, as an extension from the above stalemate question

On the other hand, there are some sets that fail without a king:

  • B(w) + B(b) (but 3 bishops should work, so long as one is on a different color)
Nick T
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  • 3 bishops do not work to force a mate - the lone color bishop is not sufficient to force the king to a corner. Four bishops works though. Four knights, but not three (I think, after trying for a while). This would best be answered by an engine which can calculate without a King. – Daniel Jul 02 '16 at 23:47

1 Answers1

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Assuming opponent has no pieces:

  1. 4 or more knights

  2. 4 or more bishops, at least 2 per side

  3. At least 2 pawns, but it also depends on where the pawns are

  4. 2 or more major pieces

  5. Queen and a minor piece

  6. Rook and 2 minor pieces

  7. 4 minor pieces

  8. Pawn+Minor pieces/2 Minor Pieces(depending on where the pawn is or a major piece

If opponent have pieces:

  1. Queen(maybe rook)+Minor Piece vs 1 pawn

  2. Queen(maybe rook)+2 pawns vs Rook+Bishop+Pawn

  3. Q+R vs Rook and bishop or 2 minor pieces(for most positions)

Ariana
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