Problem 3. All cells of the top row of a $14 \times 14$ square are filled with water, and one cell contains a bag of sand (see figure). In one move, Vasya can place bags of sand in any 3 cells not occupied by water, after which the water fills each of the cells that border the water (by side), if there is no bag of sand in that cell. The moves continue until the water can no longer fill new cells. How should Vasya act to ensure that the water fills as few cells as possible?

![](https://cdn.mathpix.com/cropped/2024_05_06_3fb59c7075fbe0bc8ea2g-2.jpg?height=346&width=345&top_left_y=376&top_left_x=558)
Solution. We will prove that no matter how Vasya acts, the water will fill at least 37 cells.

No matter how Vasya acts on the first move, after it, there will be no more than 3 bags in the second row, which means that the water will fill no fewer than 11 cells in the second row. No matter how Vasya acts on the second move, after it, in the first two rows there will be no more than 7 bags, that is, there will be no fewer than $14-7=7$ verticals without bags, through which the water will flow to the third row. Similarly, after the third move, at least 4 more cells will be filled, and after the fourth move, at least 1. In total, the water will fill no fewer than $14+11+7+4+$ $+1=37$ cells.

Vasya can achieve that the water fills exactly 37 cells by placing the bags, for example, as shown in the figure (on the first move, Vasya places the bags in the second row, on the second move - in the third row, and so on).

![](https://cdn.mathpix.com/cropped/2024_05_06_3fb59c7075fbe0bc8ea2g-2.jpg?height=346&width=345&top_left_y=1468&top_left_x=556)