7. Let $N=\{1,2, \cdots, n\}(n \geqslant 2), N$'s subsets $A_{1}, A_{2}, \cdots, A_{t}$ form a family of sets $F$. If for any $i, j \in N$, $i \neq j$ there exists a set $A_{k}$ such that $A_{k} \cap\{i, j\}$ contains exactly one element, then $F$ is called distinguishable. If each element of $N$ belongs to at least one subset $A_{i} \in F$, then $F$ is called covering. What is the minimum value of $t$, denoted as $f(n)$, such that a family of sets $F=\left\{A_{1}, A_{2}, \cdots, A_{i}\right\}$ is both distinguishable and covering?
（29th IMO Shortlist Problem）
7. Construct an $n \times t$ number table, where the number at the $i$-th row and $j$-th column is $x_{i j}=\left\{\begin{array}{l}1\left(\text { if } i \in A_{j}\right) \\ 0\left(\text { if } i \notin A_{j}\right)\end{array}(i=1,2, \cdots, n ; j=1,2, \cdots, t)\right.$. Since $F$ is a cover, each row contains at least one number equal to 1. Furthermore, since $F$ is distinguishable, no two rows in the table are identical. Because each number in a row can only be 0 or 1, and at least one must be 1, there can be at most $2^{t}-1$ distinct rows. Therefore, we have $2^{t}-1 \geqslant n$, $2^{t}>n, t>\log _{2} n$, so $t \geqslant\left[\log _{2} n\right]+1$. On the other hand, for any positive integer $n$, there exists a unique positive integer $t$ such that $2^{t}>n \geqslant$ $2^{t-1}$, i.e., $t=\left[\log _{2} n\right]+1$. Since $n \leqslant 2^{t}-1$, we can construct an $n \times t$ number table where each row consists of 0s and 1s, with at least one being 1, and no two rows are identical. Let the number at the $i$-th row and $j$-th column of the table be $x_{i j}$, and let $A_{j}$ be the set of all $i$ such that $x_{i j}=1$ $(j=1,2, \cdots, t)$. In this case, $F=\left\{A_{1}, A_{2}, \cdots, A_{1}\right\}$ is both distinguishable and a cover. In summary, the minimum value of $t$ is $f(n)=\left[\log _{2} n\right]+1$.