The 59th question: Given an integer $\mathrm{n} \geq 2$, let the set $\mathrm{N}=\{1,2, \ldots, \mathrm{n}\}$. The subsets $A_{i}(1 \leq \mathrm{i} \leq \mathrm{t})$ of set $\mathrm{N}$ form a family of sets $\mathcal{F}=\left\{A_{1}, A_{2}, \ldots, A_{t}\right\}$. If for every pair of elements $x 、 y \in N, x \neq y$, there exists a set $A_{i} \in \mathcal{F}$ such that $\left|A_{i} \cap\{x, y\}\right|=1$, then $\mathcal{F}$ is called "separable"; if every element of $N$ belongs to at least one $A_{i} \in$ $\mathcal{F}$, then $\mathcal{F}$ is called "covering". Determine the minimum value of $\mathrm{t}$ such that there exists a $\mathcal{F}=\left\{\mathrm{A}_{1}, \mathrm{~A}_{2}, \ldots, \mathrm{A}_{\mathrm{t}}\right\}$ that is both separable and covering.
The 59th question
Solution: As shown in the figure, construct a $\mathrm{t} \times \mathrm{n}$ grid table: where $\mathrm{t}$ rows represent the $\mathrm{t}$ subsets in $\mathcal{F}$, and $\mathrm{n}$ columns represent the $\mathrm{n}$ elements in $\mathrm{N}$. In the cell $(\mathrm{i}, \mathrm{j})$, fill in $\mathrm{a}_{\mathrm{i}, \mathrm{j}}=\left\{\begin{array}{ll}0, & \text { when } \mathrm{j} \notin A_{i} \\ 1, & \text { when } \mathrm{j} \in A_{i}\end{array}\right.$ It is easy to know that $\mathcal{F}$ is a covering if and only if the numbers in each column are not all $0$; $\mathcal{F}$ is separable if and only if no two columns are identical.

Since the elements in each column are not all $0$, at most $2^{t}-1$ distinct columns can be produced, hence $2^{t}-1 \geq n \Rightarrow \mathrm{t} \geq 1+\left[\log _{2} \mathrm{n}\right]$.

On the other hand, when $t=1+\left[\log _{2} n\right]$, $2^{1+\left[\log _{2} n\right]}-1 \geq n$ distinct columns with elements not all $0$ can be produced.
In summary, the minimum value of $\mathrm{t}$ that satisfies the conditions is $1+\left[\log _{2} \mathrm{n}\right]$.
Note: The analysis using a graph is as follows: Construct a bipartite graph $\mathrm{G}=\mathrm{A}+\mathrm{B}$, where $\mathrm{A}$ contains $\mathrm{t}$ points representing the $\mathrm{t}$ subsets in $\mathcal{F}$, and $\mathrm{B}$ contains $\mathrm{n}$ points representing the $\mathrm{n}$ elements in $\mathrm{N}$. If an element $\mathrm{j}$ in $\mathrm{N}$ belongs to a subset $A_{i}$ in $\mathcal{F}$, then connect an edge between the corresponding two points. It is easy to know that $\mathcal{F}$ is a covering if and only if the neighborhood of each point in $\mathrm{B}$ is not an empty set; $\mathcal{F}$ is separable if and only if the neighborhoods of any two points in $\mathrm{B}$ are different.

Thus, the condition is equivalent to the neighborhoods of the $\mathrm{n}$ points in $\mathrm{B}$ being $\mathrm{n}$ distinct non-empty subsets of $\mathrm{A}$. Since $\mathrm{A}$ has exactly $2^{t}-1$ non-empty subsets, $2^{t}-1 \geq n \Rightarrow t \geq 1+\left[\log _{2} n\right]$. And when $t=1+\left[\log _{2} n\right]$, $\mathrm{A}$ has $n$ distinct non-empty subsets. Therefore, the minimum value of $\mathrm{t}$ that satisfies the conditions is $1+\left[\log _{2} \mathrm{n}\right]$.