7.10 Let $\left.k \in N, S_{k}=|(a, b)| a, b=1,2, \cdots, k\right\}$. For two elements $(a, b)$ and $(c, d)$ in $S_{k}$, if
$$
a-c \equiv 0 \text { or } \pm 1, b-d \equiv 0 \text { or } \pm 1(\bmod k),
$$

then $(a, b)$ and $(c, d)$ are said to be indistinguishable in $S_{k}$ (for example, $(1,1)$ and $(2,5)$ are indistinguishable in $S_{5}$). Otherwise, they are distinguishable.

Consider a subset $A$ of $S_{k}$ with the following property: all elements in $A$ are pairwise distinguishable in $S_{k}$. The maximum number of elements in such a subset is denoted by $r_{k}$.
(1) Find $r_{5}$ and explain the reasoning;
(2) Find $r_{7}$ and explain the reasoning;
(3) For a general $k$, what is $r_{k}$ (no need to explain the reasoning)?
[Solution] We use $k \times k$ squares to represent the $k^2$ elements of $S_k$, naturally with the square in the $i$-th row and $j$-th column representing $(i, j)$. Thus, two elements $(a, b), (c, d)$ in $S_k$ being indistinguishable is equivalent to their corresponding squares being adjacent (including adjacent with a common edge or a common vertex), where adjacency is defined broadly, meaning the first row (column) is adjacent to the $k$-th row (column). Clearly, each square is adjacent to exactly 8 surrounding squares for $k \geqslant 3$.

For any $2 \times 2$ square, the 4 squares within it are pairwise adjacent, so the 4 corresponding elements in $S_k$ are pairwise indistinguishable. Therefore, any $2 \times 2$ square can contain at most one element of $A$.

Consider any two adjacent rows in the grid. We append another $2 \times k$ grid to the side of the $2 \times k$ grid, thus obtaining a $2 \times 2k$ grid, which contains $k$ non-overlapping $2 \times 2$ squares. Since any $2 \times 2$ square can contain at most one element of $A$, the $2 \times 2k$ grid can contain at most $k$ elements of $A$ (note the broad definition of adjacency). However, for the original $2 \times k$ grid, each square is used exactly twice in the $2 \times 2k$ grid, so the $2 \times k$ grid can contain at most $\left[\frac{k}{2}\right]$ elements of $A$, meaning any two adjacent rows can contain at most $\left[\frac{k}{2}\right]$ squares in $A$.

Using the same technique, we can prove that in $k$ rows, there can be at most $\left[\frac{k}{2}\left[\frac{k}{2}\right]\right]$ squares in $A$, so we have
$$
r_k \leqslant \left[\frac{k}{2}\left[\frac{k}{2}\right]\right].
$$

Next, we will prove that the equality holds in the above formula. According to the problem, we only need to provide examples for $k=5$ and $k=7$. We only need to give a set of non-adjacent squares that has exactly $r_k$ elements. Note that $r_5 = 5$ and $r_7 = 10$, the selected set of squares is as follows:
For general $k$, we also have $r_k = \left[\frac{k}{2}\left[\frac{k}{2}\right]\right]$.