I don't know, but I think my opponent just may be cheating in this chess game.

enter image description here

So, the natural thing to do is cheat right back! My plan is to put black pieces on the board so the white king is immediately checkmated.

However, I need to put the pieces with the lowest total value (pawn = 1, knight/bishop = 3, rook = 5, queen = 9) on the board to checkmate the king.

So, the question is:

What is the lowest total value of black pieces you can 'drop' on the board to immediately checkmate the white king here?

Some other rules:

  • Normal piece counts don't apply. Go wild with 10 pawns, 5 knights, 4 rooks, etc.

  • Legal double checks are allowed; triple checks aren't.

  • You may only drop black pieces, not additional white ones.

Good luck!

  • 1
    $\begingroup$ What does "legal double checks" mean? $\endgroup$
    – Gareth McCaughan
    Oct 16, 2018 at 20:17
  • $\begingroup$ what about the black king? $\endgroup$ Oct 16, 2018 at 20:18
  • $\begingroup$ @RomanOdaisky feel free to ignore the king, and I'm not allowing it in the puzzle $\endgroup$ Oct 16, 2018 at 20:18

2 Answers 2


I doubt this is minimal but I can do it with

1+3+9=13 "points": Qh5 supported by Pg6, and Nd5 to cover the other two squares the WK can move to.

[EDITED to add:] Better:

1+1+1+1+1+3 = 8 "points": pawns on {g,h}{6,7} and h5, plus bishop on g3.

  • $\begingroup$ and what was black’s last move? $\endgroup$ Oct 16, 2018 at 20:30
  • 2
    $\begingroup$ I don't think there are moves in this scenario: this doesn't need to be a position that can legally be reached, or anything like that. ("10 pawns", "feel free to ignore the [black] king", etc.) $\endgroup$
    – Gareth McCaughan
    Oct 16, 2018 at 20:39
  • $\begingroup$ you could move things around a bit (4Q3/4Qpp1/4Q1pp/4Q1Kp/4Q3/4Q1b1/4Q3/4Q3) though. And I think it can’t be improved (pawns can’t attack h4 so you need at least one minor piece, and no more than two pawns can attack two relevant squares). $\endgroup$ Oct 16, 2018 at 20:44
  • $\begingroup$ Yeah, that’s it. 8 is the minimum as far as I can tell. $\endgroup$ Oct 16, 2018 at 20:47
  • $\begingroup$ For what it's worth, it feels minimal to me too for the reasons Roman gives. $\endgroup$
    – Gareth McCaughan
    Oct 16, 2018 at 21:03

Expanding on all possible solutions..

8 points as previously mentioned
There are three mutually exclusive things that can change (the 7th rank, 5th rank, and g file), with two possibilities each. There are 3 binary choices, with the same number of possibilities as 111 in binary, which is 7.


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