Using only kings, and as few rooks as possible, set up a position where 30 of black's kings are checkmated.


  1. The king is in check
  2. The king has no moves out of check

Black only has kings, exactly 30 of them. White has rooks and kings, and the goal is to use as few rooks as possible.

  • White must have at least one king, but can have more than one.
  • White kings and black kings can not stand next to each other.
  • All black kings must be checkmated, stalemate is not enough.


Left: NOT valid. The king on g8 is checkmated, but the king on h8 is merely stalemated.
Right: A valid solution for 2 kings with 1 rook.

enter image description here enter image description here

Left: NOT valid. Either king can capture the rook.
Right: A valid solution for 3 kings with 2 rooks.

enter image description here enter image description here

Online board editor

  • 2
    $\begingroup$ In that case, White King on c4 should be on b4 , then it becomes a Valid Solution $\endgroup$
    – Prem
    Commented Jun 22, 2022 at 19:50
  • 1
    $\begingroup$ Even though this is a generalization of chess (as known by common rules) it shows wonderful imagination to even think of this problem! $\endgroup$ Commented Jun 22, 2022 at 20:49
  • 1
    $\begingroup$ Is it better to have 3 kings and 2 rooks, or 2 kings and 3 rooks? $\endgroup$ Commented Jun 23, 2022 at 3:11
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    $\begingroup$ @d-b Then what is the point of any question on this site, let alone its existence? $\endgroup$ Commented Jun 24, 2022 at 5:46
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    $\begingroup$ As @Prem said, the last board has more white kings than necessary. Moving the one from c4 to b4 would still get the job done with 2 total, you didn't need to add another, unless you're trying to demonstrate that sub-optimal solutions can be valid. Not sure if it's intentional, since Prem's comment pointing that out came 10 minutes before your last edit. $\endgroup$ Commented Jun 25, 2022 at 11:18

6 Answers 6


Here's a solution with 11 rooks and 1 king:

enter image description here enter image description here

kkkRkkRk/2kRkkRk/R1kkkkkk/2k2Rk1/k1k1R1k1/k4Rk1/kRkk3R/kR1kkk1K b - - 0 1 https://lichess.org/editor/kkkRkkRk/2kRkkRk/R1kkkkkk/2k2Rk1/k1k1R1k1/k4Rk1/kRkk3R/kR1kkk1K_b_-_-_0_1

  • $\begingroup$ Impressive. Two neat rook positions where distance takes care of rook not being attacked. Is this the unique solution or are there more? I presume this is the optimal and that there is no solution with 10, right? $\endgroup$ Commented Jun 23, 2022 at 7:56
  • $\begingroup$ There's at least one other solution, if the G5 king moves to H5 $\endgroup$
    – user7868
    Commented Jun 23, 2022 at 8:34
  • 1
    $\begingroup$ @user7868 nope, h6 doesn't get checkmated then. Similar issue with e1 king if you move c2 to e2. And various other moves. I don't see a move that would leave all kings checkmated with this solution. $\endgroup$ Commented Jun 23, 2022 at 8:51
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    $\begingroup$ I have encoded this problem as a SAT problem, and if my encoding isn't wrong, then 10 rooks is in fact impossible, and there are tons of 11 rook solutions. $\endgroup$
    – Mateon1
    Commented Jun 23, 2022 at 9:14
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    $\begingroup$ I left my SAT encoding version run in the background today to figure out what's the maximum number of kings you can pack into an 11-rook solution. 5 white kings is impossible, but here's one with 4 white kings: KK1kR2k/R2k1K1k/kkkk3R/kRRkk1K1/kkkkk3/1kkRRkkk/RRkkkkRk/kkk3Rk w - - 0 1 $\endgroup$
    – Mateon1
    Commented Jun 23, 2022 at 18:05

To kick things off, here is a solution with 1 white king and 14 white rooks.

enter image description here

I have a feeling this could be tweaked to remove a rook by adding some more kings, but haven't quite managed yet.

  • 4
    $\begingroup$ Good start. It is indeed possible to do it with less than 14 rooks. $\endgroup$ Commented Jun 22, 2022 at 8:20
  • $\begingroup$ It looks like black can escape and win with Kd6 to d7, assuming black can put one king in jeopardy because it still has many kings left and white can't attack back because it has only one and d7 is guarded by Ke6. $\endgroup$ Commented Jun 25, 2022 at 14:21
  • $\begingroup$ @JohnChurchill If that was the way things worked in this puzzle, some of the examples given in the problem statement also fail. $\endgroup$ Commented Jun 25, 2022 at 15:52

This just gives a lower bound on what the best solution could be.

For a black king to be in mate, all neighboring fields must be either threatened or blocked and the field where the king is must be threatened. The blocking of neighboring fields can be done in a variety of ways but threatening the king itself can only be done by a rook. A rook can move in any of the 4 directions but it can only threaten one king in every direction. Hence you need at the very least one rook per 4 kings which gives a lower bound of 8 rooks to mate all 30 kings.

At the time of writing the best solution has 11 rooks, I don't know where in the range of 8 to 11 the ideal solution is.


Solution with 12 rooks and 3 kings: https://lichess.org/editor/K1k2Rkk/2kkkRk1/1kRRkkk1/2kkkkRR/K1kkRkk1/3kRkkk/1KRk1kRk/1R1kkkRk_w_-_-_0_1 12 rook solution

There are some trivial modifications to add 2 checkmated black kings (move Rb1 to c1, remove Kb2, put black kings on a1 and a2), but I don't see a way to remove a rook.

  • $\begingroup$ Now that there's 2 solutions with the same number of rooks, feels silly to post a 3rd $\endgroup$ Commented Jun 23, 2022 at 2:13

Here is a solution with 1 king, 13 rooks.

It feels like it might be possible to remove one rook with a few adjustments, but I haven't achieved this after hours of shuffling pieces around.

Mate 30 kings with 13 rooks

For a slightly different challenge, here is a way to mate 41 black kings:

mate 41 kings with king and rooks

  • 5
    $\begingroup$ I guess this possible alternate challenge could be defined as maximizing kings instead of minimizing rooks. Extending @Jaap Scherphuis's pattern, I found one with 42 kings $\endgroup$ Commented Jun 22, 2022 at 8:29
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    $\begingroup$ @SE-stopfiringthegoodguys oups, thank you. I have edited a correct version in. I had shuffled to much trying to save one rook and posted a bad version ! $\endgroup$
    – Evargalo
    Commented Jun 23, 2022 at 9:36

Another solution with 12 rooks and 3 kings:

enter image description here

FEN: k1RR1k1K/kkkkkk2/RRkkRRk1/kkkkkk2/k1RRk2K/kkkkk1K1/RRkkR3/kkk1Rk2 w - - 0 1


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