Suppose that the chess board is not a 2D square, but a cylinder, as shown below:
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:
If it is white's move, then
Qg7 is mate.
If it is black's move, then black has to play
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
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
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?