Anozhans A.

A square is divided into $n^{2}$ equal smaller squares. It is known that a certain broken line 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?

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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_0918ecac094e52bfe73bg-19.jpg?height=500&width=552&top_left_y=1122&top_left_x=753)

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.