Given is an $n\times n$ square of dots (so $3\times3$ is a $2 \times 2$ square, which means $9$ dots: corners of the squares).
What have I already achieved and proven?
Obviously for $n=1$: The answer is one: Through one point one line is enough and need to touch all points.
For $n=2$, the answer is three.
For $n=3$, it's not five, it's four as known in the popular problem:
Then I decided to check how we got here. I drew the setups for $n=4, 5, 6, 7, 8$ where the value of segments was equivalently $6, 8, 10, 12, 14$ using different strategies to connect the dots. I was stuck there for some hours, but in the end I proved with induction that for $n > 2$ the amount of segments possible is $2(n-1)$
You took the picture above ($^2$) and just continued the diagonal one more, and always added two more lines for one higher n:
Since we arrive in the relative the same position, any $n\times n$ dots grid can be filled with $2n - 2 = 2 (n-1)$ segments, without holding off the pencil. (We originally just draw spirals, which was $2n$, the we started to draw a kind of Z/snake figure which was $2n-1$ for all $n$)
What do I still have to do?
I proved that it's possible, but not that it's the minimum. If it's the minimum, how to prove that the amount of segments needed is actually minimal, and you can't go further? Is the formula below correct (which is simply my hypothesis)? Is the current formula showing the absolute minimum at all right now?
$$ \left\{\begin{array} \\ t_1 = 1 \\ t_2 = 3 \\ \forall n > 2, n \in \mathbb{N}: t_n = 2(n-1) \end{array}\right. $$
Because I have the feeling, that for, e.g. hundred times hundred square, the square will not consist of $2(100-1) = 198$ segments, but perhaps even 195?