Bernardo can place shots along the full length of one diagonal. That will guarantee his victory. So the lowest value of $m$ is $n$.

SeanTran wrote: Bernardo can place shots along the full length of one diagonal. That will guarantee his victory. So the lowest value of $m$ is $n$. That is not true, since Bernardo wont be able to tell the difference between a vertical boat and a horizontal one.

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The lowest value for $m$ is $\left\lceil \frac{4n}{3} \right\rceil + c$, where $c$ is the remainder of $n \mod$ $3 $. Let's begin with the case $n \equiv 0\mod 3$. Let's consider the $3 \times 3$ squares in the diagonal ($\frac{n}{3} $ in total). For every one of these squares, Bernardo will choose all squares in the first row (from top to bottom) and all the squares in the last column (from left to right), except the square at the top right, four cells for every little square. Suppose that Arnaldo has chosen a set of boats, and let's observe at a square that corresponds to a position occupied by a ship. By construction, this square has a neighbor square which was also chosen by Bernaldo (neighbor meaning sharing a side). That neighboring square will help us determine wheter the boat passing through the first square is in horizontal or vertical position. Once we have determined if the boats are vertical or horizontal, it is easy to see in which columns/rows there are boats. In this way, Bernaldo will always be able to determine the position of the boats. Now let's argue why this $m$ is minimal. Suppose we have $m < \left\lceil \frac{4n}{3} \right \rceil$. Divide the square into $3 \times n$ horizontal stripes, by pigeongole principle there is at least one stripe with at most $3$ marked houses on it (and the same is true for $n \times 3$ stripes. Consider the union of these two chosen stripes. It is easy to show that it is impossible for Bernaldo to tell the difference between two vertical boats inside this union and two horizontal boats inside this union. For the cases $n \equiv 1 \mod 3$ and $n \equiv 2 \mod 3$, we cover the board with the same idea, and use the remaining $n \mod 3 + 1$ squares to fill the other columns/rows, making sure that every chosen square shares a side with at least another chosen square. I'm having trouble formalizing the minimality argument for these two remaning cases, in the next edit I will surely add them.