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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?

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    $\begingroup$ 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 '17 at 17:44
  • $\begingroup$ Can you capture a rook on an empty row, because it will be self protecting. $\endgroup$ – prog_SAHIL Dec 30 '17 at 17:57
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    $\begingroup$ "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 '17 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 '17 at 20:04
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    $\begingroup$ @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 '17 at 20:12
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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%).

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  • $\begingroup$ This feels more like a specific endgame rather than a forced variation. $\endgroup$ – prog_SAHIL Dec 30 '17 at 18:30
  • $\begingroup$ That is a very clever solution. Seems to be working well. $\endgroup$ – padawan Dec 30 '17 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 '17 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 '17 at 19:52
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    $\begingroup$ @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 '17 at 2:04
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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.

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