On a chessboard, a pawn is located on B8 and another on G8, while a knight is on B1. What is the minimum number of moves required for the knight to capture both pawns? (The pawns do not move in between.)

![](https://cdn.mathpix.com/cropped/2024_05_02_2ac192c2ab0643520972g-1.jpg?height=442&width=455&top_left_y=278&top_left_x=824)
Solution. The knight will move in such a way that it first captures the pawn on B8 or G8, and then from that position, it will capture the other pawn. The second sequence of moves is clearly independent of which pawn was captured first, as the direction of the moves can be reversed. The knight moves one or two columns and rows at a time, so at least three moves are required to get from B8 to G8. This is solvable in three moves, for example: B8, D7, F6, G8. To move from the 1st row to the 8th row, at least four moves are needed based on the above. Four moves are sufficient to capture the pawn on G8: B1, C3, D5, F6, G8. Therefore, at least 7 moves are required to capture both pawns, and this is sufficient.

Note. The knight can only capture the pawn on the dark G8 square with its first four moves, as the knight lands on a square of the same color as the one it started from after an even number of moves. Capturing the pawn on B8 would require at least 5 moves, but this does not give the minimum number of moves.