Let $A$ be the following.

A square is divided into $n^{2}$ equal smaller squares. It is known about a certain broken line that it passes through the centers of all the smaller squares (the broken line can intersect itself). What is the minimum number of segments in this broken line?

#
In the figure, an example of such an eight-segment broken line for $n=5$ is shown. It consists of a known traversal of nine points by a four-segment broken line and an unwinding spiral.

![](https://cdn.mathpix.com/cropped/2024_05_06_f7e07b07084714df1adcg-32.jpg?height=492&width=546&top_left_y=1132&top_left_x=756)

Obviously, by continuing to unwind the spiral, we can obtain an example of a $(2n-2)$-segment broken line for any $n$. Let's prove that fewer segments are not sufficient. Suppose that through the centers of the squares, there are $l$ horizontal and $m$ vertical segments.

If $l \geq n$, then (since horizontal segments cannot be adjacent) the number of segments is at least $n + (n - 1) = 2n - 1$.

If $l = n - 1$, then there remains a horizontal row of $n$ points not lying on horizontal segments. At least $n$ non-horizontal segments are needed to "cover" these points, and the number of segments is again at least $2n - 1$.

Finally, let $l, m \leq n - 2$. Remove all points (centers of squares) lying on lines passing through horizontal and vertical segments. A "rectangle" of $(n-l)(n-m)$ points remains. On its "boundary" lie $2(n-l-1) + 2(n-m-1)$ points. Any inclined segment of the broken line contains no more than two "boundary" points, so there are at least $2n - l - m - 2$ such segments. In total, the broken line has at least $l + m + (2n - l - m - 2) = 2n - 2$ segments.

## Answer

$2n - 2$ segments.