Suppose that the chess board is not a 2D square, but a cylinder, as shown below:

enter image description here

In that case, both players still can make queens, but one king and one queen is not enough to checkmate the opponent, simply because there is no corner to drive the opponent king. Consider the below position in a square chessboard:

![enter image description here

If it is white's move, then Qg7 is mate. If it is black's move, then black has to play Kg8, and Qg7 is still checkmate.

This mate exploit the fact that black king could not escape from the corner the move before. But if the board is cylindrical, then black has the move Ka8!.

On the other hand, if white had a pawn on b7, then after Qg7, would have been a checkmate because unlike 2D board, white queen also controls the a7 square.

enter image description here

So, the question is:

What is the minimum total value of the pieces that you can deliver a forced checkmate as white, in a cylindrical chessboard to a single black king?

Note that forced means you cannot assume a position, but only a material balance.

The values for the pieces:

Pawn: 1  
Bishop: 3  
Knight: 3  
Rook: 5  
Queen: 9

If a pawn promotes to any piece, then it means the value of the pawn is changed to that piece. The score is calculated by the final position on the board.

I have modified the value of the knight, because it no longer is a passive piece at the edge of the board. On the other hand, bishop still controls the same amount of squares.

As @Sleafar pointed out in the comments, you can actually deliver checkmate with a queen and a king in a cylindrical board. But it is still 9. What is the minimum?

  • 1
    $\begingroup$ On the other hand, bishop still controls the same amount of squares. On a traditional board, a bishop in a corner controls 7 squares (excluding its own); on the cylindrical board, it controls 13. $\endgroup$
    – Apep
    Dec 30, 2017 at 17:44
  • $\begingroup$ Can you capture a rook on an empty row, because it will be self protecting. $\endgroup$
    – prog_SAHIL
    Dec 30, 2017 at 17:57
  • 1
    $\begingroup$ "In that case, both players still can make queens, but one king and one queen is not enough to checkmate the opponent, simply because there is no corner to drive the opponent king." - This is simply a wrong assumption, you don't need to corner the king, a simple back-rank checkmate can be forced with a queen. E.g. in your first example 1. Qb7 Kg8 2. Qa7 Kf8 3. Qh8# $\endgroup$
    – Sleafar
    Dec 30, 2017 at 19:27
  • $\begingroup$ @Sleafar But you cannot drive the king back rank. This is true because the king is already at the back rank. $\endgroup$
    – padawan
    Dec 30, 2017 at 20:04
  • 2
    $\begingroup$ @padawan You can drive the king to the back-rank in exactly the same way as I showed above. Just follow the black king with your queen until opposition is reached, then check with the queen, rinse and repeat. $\endgroup$
    – Sleafar
    Dec 30, 2017 at 20:12

4 Answers 4


This question needs a lot of work. For example, a single queen is quite enough to force the opposing king to one of the two edges and then force a checkmate. (From the sample ”impossible” position given, Qb7 forces a mate in two.)

The worst problem is the scoring, though. I claim I can mate the king from any reasonable starting position with

five points. I start with a bishop and two pawns.

As to the strategy,

barring pathological cases, I can easily promote both pawns. I choose a queen and a rook. Then I use the overwhelming firepower to forcingly sacrifice the bishop for no gain.

Then I easily force the king to one of the two sides. (I chose the top side in the example.)

with the rook guarding the seventh rank and the king on the sixth, I use the queen on the fifth rank to herd the opposing king to this position with black to move:
enter image description here

Then black only has one possible move, after which

I sacrifice the queen, forcing the king into opposition, and mate with the rook on the eighth rank. This counts as five, since the problem explicitly states that ”the score is calculated by the final position on the board

I know this isn’t what OP intended when he wrote out the rules for the puzzle, so please regard this more as a ”pointing out the glaring loophole” thing rather than as a serious answer.


It's possible to

checkmate the majority of positions with a rook, pawn and king (6 points), without promoting the pawn. If pawn promotions are allowed, some king + pawn vs. king endgames can be won because king + queen vs. king is still a win; e.g. from the first diagram in the question, play 1. Qb7 Kg8 2. Qg7#.

Assume you're White

and the Black king is already on the back rank, with your rook on the seventh. Move the king on the sixth rank, and advance the pawn to the seventh rank (but make sure it's not blocked by the opposing king, as that would be stalemate; use the rook to drive the Black king away). This limits the Black king's space to five fields. The rook can freely move between the squares next to the pawn if necessary to avoid being attacked.

Without loss of generalization (it's a cylindrical board), assume the pawn is at g7. Move your king to c6; if Black plays Kc8, you can give mate on the back rank. If Black plays Ka8 or Ke8, play Kb6 resp. Kd6, and it's mate on the next move. If Black plays Kb8 or Kd8, play a waiting move with your rook.

If the King is not yet on the back rank, but still higher up the board than the pawn (as Sleafar notes), you can use the same procedure to drive the King a rank up the board. That will make this endgame winnable at least than 50% of the cases (and, given the nature of chess games, where kings and pawns tend to stay close to their home ranks, way more than 50%).

  • $\begingroup$ This feels more like a specific endgame rather than a forced variation. $\endgroup$
    – prog_SAHIL
    Dec 30, 2017 at 18:30
  • $\begingroup$ That is a very clever solution. Seems to be working well. $\endgroup$
    – padawan
    Dec 30, 2017 at 19:34
  • $\begingroup$ Driving the king with rook and king to the back-rank "as normal" works because the board is not cylindrical. On a cylindrical board you cannot gain opposition with your king against the other king, because it can simply run away further left or right. $\endgroup$
    – Sleafar
    Dec 30, 2017 at 19:45
  • $\begingroup$ @Sleafar indeed, you're right. It seems that (as long as you can keep the Black king on the third rank or above) repeated application of the final mating procedure will work to drive the king a rank up. So the material balance (+6) is right, but the procedure needs to be altered slightly. I'll edit my answer and add some diagrams. $\endgroup$
    – Glorfindel
    Dec 30, 2017 at 19:52
  • 1
    $\begingroup$ @Glorfindel It is a specific endgame (king, rook + pawn vs. lone king), but this strategy should work for all positions with these pieces, excluding a handful where Black can capture one of them on the first or second move. That's not forced then. $\endgroup$
    – prog_SAHIL
    Dec 31, 2017 at 2:04

Edit: This solution doesn't seem to work I'll leave it to show how having a cylindrical board can affect endgame outcomes - sometimes counterintuitively !

Another solution with

6 points

is to mate with

two bishops.


The usual mating procedure with KBB vs K in standard chess is also effective in cylindric chess, because the range of the bishops isn't stopped by the sides of the board, and prevents lateral flights by the bK. Said otherwise, you don't need a corner to mate with KBB (unlike with KR or with KBN). Edit: well, you do!

An example:

KBBK mate

Note that

during this whole sequence the bK has never been allowed to cross the "outer border" between the a- and h- files. Notably, the wB prevented ...Ka8-h8 in the final moves from either d4 or b6. Edit: however, one move before mate, Black can play Ka8-h7 instead of Ka8-b8. This possible diagonal flight spoils any try to mate with KBB vs K.

  • 1
    $\begingroup$ Excellent solution that works from any legal position, white to move. The bishop/pawn solutions do not work if the black king is on the first or second rank. And excellent moving diagram to show it works. $\endgroup$ Aug 9, 2019 at 16:43
  • 1
    $\begingroup$ At some point you are in the position k1B5/8/1BK5/8/8/8/8/8 w - - and Black plays Kb8, but what if they go Kh7? $\endgroup$ Aug 10, 2019 at 23:22
  • $\begingroup$ @ArnaudMortier You are right, good catch, I had missed that possibility. It can be corrected easily with Be6-c4 instead of Be6-c8 on the previous move - I'll edit the answer once I am on a computer again, dealing with animated diagrams with my smartphone is too messy. Edit: oh, actually this is also a problem one move before mate, and a more serious one. I have to think and reconsider my answer. $\endgroup$
    – Evargalo
    Aug 11, 2019 at 11:08

With a King, Pawn and Rook (6 points), it is straightforward to force the black King back toward the final rank.

The key is to have the rook protected by the pawn, with the black king on a higher rank. Eg.:

enter image description here

From here, the white king chases the black king to get opposite it (same column, 2 ranks down), and with the black king at least 2 columns away from the rook, and white to move.

enter image description here

Since black can't run away horizontally forever, the white king can definitely catch it. Also, white can adjust a tempo if necessary by moving down a rank and back up. So white can definitely force this position.

From here, white white advances the rook one rank, forcing the black king to retreat. White then advances the pawn on the next move to protect the rook and repeat the process.

If the black king retreats before being forced back, eg.:

enter image description here

then the white king can advance to force the black king to stay back. Then advance the pawn and rook (pawn first if the black king threatens a rook capture if the rook advances, otherwise rook first).

Eventually, the black king is on the 8th rank, the white rook on the seventh, protected by the pawn. The white king chases the black king to be opposite it, then advances the rook for mate.

Thus, white can force a mate from any starting position where the black king starts on at least rank 3 and is far enough from the starting positions of the white pieces that it cannot immediately capture one of them.

I don't think it's possible for a white to force the black king off the first or second rank, but this solution is generic enough.

If you need a solution that works from any position, then replace the pawn with a bishop and white can force a win from any position (where white can't make an immediate capture). That would be a total of 8 points, still cheaper than using a queen.

  • $\begingroup$ Hi! In the future, try to hide your solutions with spoiler tags, set off by ">!". $\endgroup$
    – Cloudy7
    Jul 25, 2019 at 16:49
  • $\begingroup$ I was going to, but figured the question was over a year and a half old, maybe it wasn't worth the bother? Is it typical to use spoiler tags even if the question is old? $\endgroup$ Jul 29, 2019 at 15:49

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