Clearly not a practical question since the king can and should be used in the endgame, but I was curious if anyone has done a comprehensive analysis of which combinations of pieces (including some unrealistic combinations) can force a checkmate against a lone king in a pawnless endgame without being aided by their own king. Let's just pretend the attacking king doesn't exist (or maybe, as a looser restriction, tuck it in a remote corner where it might be allowed to make waiting moves).

There are some easy combinations that can force a checkmate:

  • queen and any other piece
  • two rooks

On the other hand, there are these classic endgames where it is possible to mate with the help of the king, but not without it:

  • queen
  • rook
  • two bishops
  • bishop and knight

It seems to me that the following combinations of pieces can force a checkmate:

  • rook and two bishops
  • two bishops and two knights
  • four bishops (two pairs!)

And the following don't:

  • rook and bishop
  • rook and knight (you can mate, but can you force it?)

But there are many other combinations I haven't looked at in detail yet:

  • rook and two knights
  • rook and knight and bishop
  • two bishops and knight
  • two knights and bishop
  • three knights
  • three bishops
  • four knights
  • ...

2 Answers 2



queen and any other piece

The queen can on its own push the king to the edge of the board by following it a knight-distance away and making use of zugzwang. Once on the edge, it is easy to bring the second piece and mate with the queen in front of the king. The only exception would be if the second piece is a pawn on the next to last row

Example for a knight (similar for any other piece).

3k4/3Q4/8/4N3/8/8/8/4K3 w - - 0 1

two rooks

Well known lawnmower mate.

rook and two bishops

Easy mate. Two bishops cover the escape route of the king while the rook checks from the side. Sample game to demonstrate the method:

8/8/8/8/3k4/8/6RB/6KB w - - 0 1

1. Re2 Kd3 2. Re8 Kd4 3. Bf4 Kd3 4. Bf3 Kd4 5. Rd8+ Kc4 6. Be3 Kc3 7. Be4 Kc4 8. Rc8+ Kb4 9. Bd5 Kb5 10. Rb8+ Ka5 11. Bc5 Ka6 12. Bc6 Ka5 13. Ra8# *

two (opposite colored) bishops and two knights

Easy mate. Push the king to the edge of the board where the two bishops can jail the king on two squares. Then take those two squares with the two knights.

[FEN "8/8/8/8/8/3k3K/6NB/6NB w - - 0 1"]

1. Nh4 Kd4 2. Bf4 Kc4 3. Bf3 Kb4 4. Be5 Kb5 5. Be4 Kc4 6. Ngf3 Kb4 7. Nd2 Ka4 8. Bd6 Kb5 9. Bd5 Ka6 10. Bc6 Kb6 11. Bd7 Kb7 12. Nc4 Ka6 13. Bc6 Ka7 14. Bc7 Ka6 15. Nf5 Ka7 16. Nfd6 Ka6 17. Ne5 Ka7 18. Nd3 Ka6 19. Nc5+ Ka7 20. Nc8#

Two pairs of (opposite colored) bishops

Very easy mate. Four bishops cover lots of squares and it is even possible to mate the king in the middle of the board.

[FEN "8/8/8/8/8/3k3K/6BB/6BB w - - 0 1"]

1. Bf4 Kc4 2. Be4 Kc3 3. Bge3 Kc4 4. Bhf3 Kb4 5. Be2 Ka3 6. Be5 Ka4 7. Bc5 Kb3 8. Bb5 Ka2 9. Bc4#

rook + bishop + knight

Easy mate. Similarly to rook and two bishops you can cover the escape squares while the rook checks from the side.

[FEN "8/8/8/8/3k4/7K/6NR/7B w - - 0 1"]

1. Nh4 Kc3 2. Nf5 Kb3 3. Rd2 Kc4 4. Be4 Kb4 5. Rc2 Kb3 6. Bd3 Kb4 7. Nd4 Ka5 8. Rb2 Ka4 9. Bc4 Ka3 10. Rb6 Ka4 11. Nc2 Ka5 12. Ra6# 

rook and two knights

Easy mate. Similar to the king+rook mate, you can create a box with one knight protecting the rook. You have the second knight free to take further squares from the king and thereby make the box smaller. Mate is similar to rook + two bishops or rook+bishop+knight with the rook checking from the side and the knights taking away the escape squares.

[FEN "8/8/8/8/3k4/8/6NR/6NK w - - 0 1"]

1. Rh3 Ke5 2. Rh4 Kd6 3. Nf3 Kc5 4. Rd4 Kb5 5. Ne3 Kc6 6. Rd5 Kc7 7. Nd4 Kb6 8. Nc4+ Kc7 9. Rd6 Kc8 10. Nb5 Kb8 11. Rd7 Kc8 12. Rh7 Kd8 13. Ne5 Ke8 14. Nd6+ Kf8 15. Nf5 Kg8 16. Rh6 Kf8 17. Rh8#

four knights

Easy mate. Keeping the knights together you take lots of squares and it is easy to push back the king and mate him.

[FEN "8/8/8/8/3k4/7N/6NN/6NK w - - 0 1"]

1. Nf2 Kc4 2. Nhf3 Kb4 3. Nf4 Kc4 4. Nge2 Kc5 5. Nd2 Kd6 6. N2d3 Ke7 7. Ne4 Kd7 8. Nd4 Ke8 9. Ne5 Kd8 10. Nd5 Ke8 11. Nd6+ Kf8 12. Ne6+ Kg8 13. Nf6+ Kh8 14. Ndf7#


  • queen or
  • rook or
  • two bishops or
  • two knights or
  • bishop and knight or
  • rook and bishop

There is not even a theoretical mating position for these.

rook and knight

There is a theoretical mate (Arabian mate), but it cannot be enforced. For an analysis see this question

two (opposite colored) bishops and knight

This is not obvious. Let's just look at the potential mating positions. You can easily convince yourself that the only way to mate is if the king is in the corner (let's say in a8, similarly for other corners) and if the mate is done by a bishop (along the a8-h1 diagonal) while the other bishop and the knight cover the escape squares b8 and a7. I can see three fundamentally different mates. (other mates are mirror symmetric or differ only in the distance of the bishops so are essentially the same).

k7/B7/8/1N6/4B3/8/8/8 w - - 0 1


k7/8/N7/8/3BB3/8/8/8 w - - 0 1


k7/3N4/8/8/3BB3/8/8/8 w - - 0 1

Doing a small retro-analysis, you see that in all cases the last move was Be4+ and that bishop came from somewhere along the b1-h7 diagonal.

What was black's last move?

  • If the king came from a7 or b8, it could as well have gone to b7 in the last move (note that b7 was not covered by any piece since the light squared bishop came from the h7-b1 diagonal). If it went instead of a8 to b7 there was no mate possible.
  • If the king came from b7 in the last move, it was not forced to move to a8 either since at least one of the squares c8 or a6 were available as well.

So in summary, there is no forced mate with two bishops and a knight.

three bishops

Similar argument as with two bishops and a knight. You easily convince yourself that (theoretical) mate is only possible if the king is in the corner of the color of the single bishop. If there are two dark squared bishops and a light squared bishop, there are essentially two different (up to symmetry) mating positions on a8:

k7/B7/8/8/3BB3/8/8/8 w - - 0 1


k7/8/8/8/3BBB2/8/8/8 w - - 0 1

In either case, the last white move was Be4 and that bishop came from somewhere along the b1-h7 diagonal.

If the king was before on a7 or b8, it could have gone to b7. If the king was before on b7 it could have gone to c8 or a6. In any case there is no way to force the king to the corner and mate him.

Impossible but no analytical proof yet

  • two knights and bishop
  • three knights

For a proof by computer analysis see the answer below by Vaclav Kotesovec.


See my book Fairy chess endings on an n x n chessboard (2017), chapter "Endings without the white king", p. 592


two knights and bishop against bare king, see p. 685. The ending is won only on a 7x7 chessboard, on boards 8x8 and greater the ending is drawn.

three knights against bare king, see p. 706. The ending is drawn on a board of any size. Mate can be forced only if the Black king is already trapped in a corner of the board.

  • Is the prove by computer analysis? Jul 14, 2020 at 16:04
  • Yes, all results in the cited book are published on the basis of computer analysis. Jul 15, 2020 at 6:20
  • Welcome to Chess Stack Exchange! Jul 21, 2020 at 2:39

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