3. Given $n^{2}$ points arranged in the plane in the form of a square grid $n \times n$. A broken line of length $l$ is the union of closed segments $A_{0} A_{1}$, $A_{1} A_{2}$, $A_{2} A_{3}, \ldots, A_{l-1} A_{l}$ in that plane. Determine the smallest natural number $l$ for which there exists a broken line of length $l$ on which all $n^{2}$ observed points lie.
3. Let the desired value of the number $l$ be denoted by $l_{n}$. Trivially, $l_{1}=1$ and $l_{2}=3$. The examples in the figure show that $l_{3} \leqslant 4$ and, in general, $l_{n} \leqslant 2 n-2$ for $n \geqslant 3$.

Assume that the broken line $\mathcal{C}$ of length $l$ covers all $n^{2}$ points. Let there be $a$ horizontal and $b$ vertical segments among the segments of the line $\mathcal{C}$. The remaining $l-a-b$ segments of the line $\mathcal{C}$

![](https://cdn.mathpix.com/cropped/2024_06_04_cfce5975f05a18722b2bg-3.jpg?height=205&width=169&top_left_y=420&top_left_x=1109)
are bent by horizontal and vertical

![](https://cdn.mathpix.com/cropped/2024_06_04_cfce5975f05a18722b2bg-3.jpg?height=354&width=323&top_left_y=366&top_left_x=1415)
segments to form a rectangular grid $S$ of dimensions $(n-a) \times(n-b)$.

- If $a \geqslant n$ (analogously for $b \geqslant n$), there is no grid $S$, but at least $n-1$ segments are needed to connect $n$ horizontal segments, so $l \geqslant 2 n-1$.
- If $a=n-1$ (analogously for $b=n-1$), the grid $S$ consists of $n-b$ points, and each diagonal segment covers at most one, so we need at least $n-b$ more segments, which gives a total of $l \geqslant(n-1)+b+(n-b)=2 n-1$ segments.
- If $a<n-2$ and $b<n-2$, the convex hull of the grid $S$ consists of $2(2 n-a-b-2)$ points, and each diagonal segment covers at most two, so we need at least $2 n-a-b-2$ more segments. Thus, in total, we have $l \geqslant a+b+(2 n-a-b-2)=2 n-2$ segments.

Therefore, $l_{n}=2 n-2$ for all $n \geqslant 3$.