A communications network consisting of some terminals is called a [i]$3$-connector[/i] if among any three terminals, some two of them can directly communicate with each other. A communications network contains a [i]windmill[/i] with $n$ blades if there exist $n$ pairs of terminals $\{x_{1},y_{1}\},\{x_{2},y_{2}\},\ldots,\{x_{n},y_{n}\}$ such that each $x_{i}$ can directly communicate with the corresponding $y_{i}$ and there is a [i]hub[/i] terminal that can directly communicate with each of the $2n$ terminals $x_{1}, y_{1},\ldots,x_{n}, y_{n}$ . Determine the minimum value of $f (n)$, in terms of $n$, such that a $3$ -connector with $f (n)$ terminals always contains a windmill with $n$ blades.
1. **Base Case: \( n = 1 \)**
   - We need to show that \( f(1) = 6 \).
   - Consider a graph with 6 vertices. Take any vertex \( A \).
   - There are two cases:
     1. \( A \) has at least 3 neighbors.
     2. \( A \) has at least 3 non-neighbors.
   - In the first case, among the 3 neighbors, at least two must be connected (since the graph is a 3-connector), forming a triangle with \( A \).
   - In the second case, the 3 non-neighbors must form a complete subgraph (since the graph is a 3-connector), forming a triangle.
   - Therefore, any 3-connector with 6 vertices contains a triangle, which is a windmill with 1 blade.
   - Hence, \( f(1) = 6 \).

2. **General Case: \( n > 1 \)**
   - We need to show that \( f(n) = 4n + 1 \).
   - **Step 1: Show \( f(n) > 4n \)**
     - Consider a graph consisting of two disjoint \( K_{2n} \) (complete graphs with \( 2n \) vertices each).
     - This graph is a 3-connector because any three vertices will either be in the same \( K_{2n} \) or in different \( K_{2n} \)'s, ensuring at least one edge.
     - However, this graph does not contain a windmill with \( n \) blades because it does not have a connected subgraph with \( 2n + 1 \) vertices.
     - Therefore, \( f(n) > 4n \).

   - **Step 2: Show \( f(n) \leq 4n + 1 \)**
     - Consider a 3-connector with \( 4n + 1 \) vertices. Take any vertex \( A \).
     - Let \( V_A \) be the set of neighbors of \( A \) and \( NV_A \) be the set of non-neighbors of \( A \).
     - Since the graph is a 3-connector, \( NV_A \) must form a complete subgraph.
     - If \( |NV_A| \geq 2n + 1 \), then \( NV_A \) contains a windmill with \( n \) blades.
     - If \( |NV_A| \leq 2n \), then \( |V_A| \geq 2n + 1 \).
     - Consider the vertices in \( V_A \). Since the graph is a 3-connector, among any three vertices in \( V_A \), at least two must be connected.
     - We can iteratively form edges among the vertices in \( V_A \) until we have \( n \) pairs, forming a windmill with \( A \) as the hub.
     - Therefore, \( f(n) \leq 4n + 1 \).

   - Combining both steps, we conclude that \( f(n) = 4n + 1 \).

The final answer is \( \boxed{ f(n) = 4n + 1 } \).