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What are the minimum sets of pieces required for checkmate when the opponent has no other piece that the king?

5
  • King + Bishop + Knight
  • King + Bishop + Bishop
  • King + Rook
  • King + Queen
  • and don't forget the unlikely King + Knight + Knight + Knight

Assuming the lone King is attempting to avoid checkmate.

  • 1
    King + Knight + Knight + Knight? – bof Dec 17 '15 at 10:13
  • @bof I doinked around with Nalimov endgame tables and could not find a case where three knights wasn't enough. Editing. – Tony Ennis Dec 17 '15 at 23:40
  • I never studied K+3N vs. K (and never saw it in a game!) bit I dimly recall that it was a win according to Fine's Basic Chess Endings. – bof Dec 17 '15 at 23:51
2

As simple as a Knight or even a pawn, if you consider the fact that your opponent's pieces can block the escape squares too.

Example:

W: Kf1, Nf2

B: Kh1, h2.

The Assume that white played Nf2# in the last move. So, white has just used one knight to checkmate black king.

  • The question specifies that the opponent has only the king left. – BlindKungFuMaster Dec 18 '15 at 9:23
1
There are many answers to your question, 

1) A rook 
2) Any 2 minor pieces (Minor pieces:  Bishop, Night)
3) A Queen
4) 1 pawn :)
  • 1
    All of those answer should include king. Also you cannot checkmate with only two knights and a king. (your 2nd case fails in that condition). – Saeed Amiri Dec 16 '15 at 15:28
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    @SaeedAmiri No, you can checkmate with two knights, it's possible just not forced. Your statement is therefore false. – SmallChess Dec 17 '15 at 0:58
  • @StudentT, According to the question : "What are the minimum sets of pieces required", it means we are talking in the context of force mate. Also if you want interpret it like this (somehow funny), I should inform you it's even impossible to checkmate with two knights (as you wrote), they cannot cover required squares. You need at least one another piece to checkmate your opponent (e.g king). – Saeed Amiri Dec 17 '15 at 10:46
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    If the 2 minor pieces are bishops, they must be on opposite color squares. – bof Dec 17 '15 at 23:57

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