$7 \cdot 111$ Let $S=\{1,2, \cdots, n\}, n \geqslant 2, F=\left\{A_{1}, A_{2}, \cdots, A_{t}\right\}$, where $A_{i} \subset S, i=1,2, \cdots, t$. If for any $x, y \in S$, there exists $A_{i} \in F$, such that $\mid A_{i} \cap$ $\{x, y\} \mid=1$, then $F$ is said to distinguish $S$. If $S \subset U A_{i}$, then $F$ is said to cover $S$. Given that $F$ both distinguishes and covers $S$, find the minimum value of $t$, denoted as $f(n)$.
［Solution 1］Let $n=2^{r}$ and define
$$
\begin{aligned}
A_{1}= & \left\{1,2, \cdots, 2^{r-1}\right\}, \\
A_{2}= & \left\{1,2, \cdots, 2^{r-2}, 2^{r-1}+1,2^{r-1}+2, \cdots, 2^{r-1}+2^{r-2}\right\}, \\
A_{3}= & \left\{1,2, \cdots, 2^{r-3}, 2^{r-2}+1,2^{r-2}+2, \cdots, 2^{r-2}+2^{r-3}, 2^{r-1}+1,2^{r-1}\right. \\
& +2, \cdots, 2^{r-1}+2^{r-3}, 2^{r-1}+2^{r-2}+1,2^{r-1}+2^{r-2}+2, \cdots, 2^{r-1} \\
& \left.+2^{r-2}+2^{r-3}\right\},
\end{aligned}
$$
$$
\begin{array}{l}
\vdots \\
A_{r-1}=\left\{1,2,5,6,9,10, \cdots, 2^{t}-3,2^{t}-2\right\}, \\
A_{r}=\left\{1,3,5,7, \cdots, 2^{t}-3,2^{t}-1\right\} .
\end{array}
$$

It is easy to see that $F=\left\{A_{1}, A_{2}, \cdots, A_{r}\right\}$ distinguishes $S$ and $U_{i=1}^{r} A_{i}=\left\{1,2, \cdots, 2^{r}-1\right\}=\{1$, $2, \cdots, n-1\}$. Clearly, adding just one more set $A_{0}=\{n\}$ to $F$ will cover $S$. Therefore, we have
$$
f(n) \leqslant r+1=\left[\log _{2} n\right]+1 .
$$

When $2^{r-1}+1 \leqslant n \leqslant 2^{r}-1$, $F$ itself (possibly intersected with $S$) is both distinguishing and covering $S$, so (1) still holds.
On the other hand, we use mathematical induction to prove the reverse inequality of (1):
$$
f(n) \geqslant r+1 \text {, when } 2^{r} \leqslant n \leqslant 2^{r+1}-1
$$

By $f(2)=f(3)=2$, we know that (2) holds for $r=1$. Assume (2) holds for $r=k$. When $r=$ $k+1$, let $F=\left\{A_{0}, A_{1}, \cdots, A_{t}\right\}$ be both distinguishing and covering $S$. Let $A^{\prime}{ }_{0}=S-A_{0}$, then $\left|A_{0}\right|+\left|A^{\prime}{ }_{0}\right|=n$, without loss of generality, assume $\left|A_{0}\right| \geqslant\left[\frac{n}{2}\right]$. At this time, $\left\{A_{1} \cap A_{0}, A_{2} \cap\right.$ $\left.A_{0}, \cdots, A_{t} \cap A_{0}\right\}$ is both distinguishing and covering $A_{0}$. Therefore, by the induction hypothesis, $t \geqslant$ $f\left(\left[\frac{n}{2}\right]\right) \geqslant k+1$. Thus, (2) also holds for $r=k+1$.
In summary, $f(n)=\left[\log _{2} n\right]+1$.
[Solution 2] Let $F=\left\{A_{1}, A_{2}, \cdots, A_{t}\right\}$ be both distinguishing and covering $S$. Consider the relation table of the subsets $A_{i}$ of $S$ and the elements
where
$$
x_{i j}=\left\{\begin{array}{l}
1, \text { when } i \in A_{j}, \\
0, \text { when } i \notin A_{j},
\end{array} \quad i=1,2, \cdots, n, j=1,2, \cdots, t .\right.
$$

Since $F$ is covering, each element $i$ belongs to at least one $A_{j}$, i.e., each row in the table is not all 0; since $F$ is distinguishing, no two rows in the table are identical. Since there are $2^{t}-1$ different non-zero vectors of $t$ components, each of which is 0 or 1, we have
$$
n \leqslant 2^{t}-1 \text {. }
$$

Thus, we get
$$
f(n) \geqslant\left[\log _{2} n\right]+1 .
$$

On the other hand, for $n \in N$, take $t \in N$ such that $2^{t-1} \leqslant n \leqslant 2^{t}-1$ and take $n$ different non-zero $t$-vectors:
$$
\left(x_{i 1}, x_{i 2}, \cdots, x_{i t}\right), i=1,2, \cdots, n,
$$

where $x_{i j}$ are all 0 or 1. Define the subset $A_{j} \subset S$ as follows: for each $i \in S$, $i \in A_{j}$ if and only if $x_{i j}=1$, $j=1,2, \cdots, t$. It is easy to verify that $F=\left\{A_{1}, A_{2}, \cdots, A_{t}\right\}$ is both distinguishing and covering $S=\{1,2, \cdots, n\}$. Therefore, we get
$$
f(n) \leqslant t=\left[\log _{2} n\right]+1 .
$$

From (3) and (4), we have $f(n)=\left[\log _{2} n\right]+1$.