11) A communication network composed of some terminals is called "triconnected" if among any three terminals, at least two are directly connected. We also refer to a communication network that satisfies the following conditions as a "windmill" with $n$ "blades": $n$ pairs of terminals $\left\{A_{i}, B_{i}\right\}$ where $A_{i}$ and $B_{i}$ are directly connected, and there exists a terminal $O$ that is connected to both $A_{i}$ and $B_{i}$ $(i=1,2, \cdots, n)$. If any "triconnected" communication network contains a "windmill" with $n$ "blades" and has the minimum number of terminals denoted as $f(n)$, find the value of $f(n)$.
Solve $f(n)=\left\{\begin{array}{ll}6, & n=1, \\ 4 n+1, & n \geqslant 2 .\end{array}\right.$
Let each point in graph $G$ represent a terminal. If two terminals are directly connected, a red line is drawn; otherwise, a blue line is drawn. According to the problem, there are no blue triangles in $G$. Let $|G|=m$.
(I) When $n=1$, a "windmill" with 1 "blade" is a red triangle.
(i) By the Ramsey number $R(3,3)=6$, when $m=6$, there must exist a monochromatic triangle. Since there are no blue triangles in $G$, the monochromatic triangle must be a red triangle, i.e., $f(1) \leqslant 6$.
(ii) When $m=5$, a counterexample can be given: draw the five edges of a regular pentagon in red and the five diagonals in blue. This graph is "3-connected" and does not contain a red triangle, hence $f(1) \geqslant 6$.
From (i) and (ii), we get $f(1)=6$.
(II) When $n \geqslant 2$.
(i) If $m=4 n$, a counterexample can be given: $A_{1}, A_{2}, \cdots, A_{2 n}$ are all connected by red lines, and $A_{2 n+1}, A_{2 n+2}, \cdots, A_{4 n}$ are all connected by red lines, and the rest are connected by blue lines. This graph is "3-connected" and does not contain a red triangle, hence $f(n) \geqslant 4 n+1$.
(ii) If $m=4 n+1$, we will prove that $G$ must contain a "windmill" with $n$ "blades".
By contradiction, assume that $G$ does not contain a "windmill" with $n$ "blades". Take any point $A_{0}$ in $G$, and let $B$ be the set of points connected to $A_{0}$ by red lines, and $C$ be the set of points connected to $A_{0}$ by blue lines.
If $|B| \geqslant 2 n+1$, let $B=\left\{A_{1}, A_{2}, \cdots, A_{2 n+1}\right\}$. Then among $A_{1}, A_{2}, A_{3}$, there must be two points connected by a red line. Without loss of generality, assume $A_{1} A_{2}$ is a red line. Then among $A_{3}, A_{4}, A_{5}$, there must be two points connected by a red line. Without loss of generality, assume $A_{3} A_{4}$ is a red line. Continue this process to find $A_{1} A_{2}$, $A_{3} A_{4}, \cdots, A_{2 n-1} A_{2 n}$, all of which are red lines. Together with $A_{0}$, they form a "windmill" with $n$ "blades", a contradiction!
If $|C| \geqslant 2 n+1$, if there are two points in $C$ connected by a blue line, then they and $A_{0}$ form a blue triangle, a contradiction! If any two points in $C$ are connected by a red line, then the $2 n$ points in $C$ and $A_{0}$ together form a red complete $2 n+1$-sided polygon $K_{2 n+1}$. In this red complete $2 n+1$-sided polygon $K_{2 n+1}$, there is clearly a "windmill" with $n$ "blades", a contradiction!
Since $|B|+|C|=4 n$, we have $|B|=|C|=2 n$, and the $2 n$ points in $C$ form a red complete $2 n$-sided polygon $K_{2 n}$. If there is a point $B_{0}$ in $B$ connected to two points $C_{1}$ and $C_{2}$ in $C$ by red lines, let $C=\left\{C_{1}, C_{2}, \cdots, C_{2 n}\right\}$. Then $C_{1}$ and $\left\{B_{0}, C_{2}\right\}, \left\{C_{3}, C_{4}\right\}, \cdots, \left\{C_{2 n-1}, C_{2 n}\right\}$ form a "windmill" with $n$ "blades", a contradiction! Therefore, any point in $B$ is connected to at most one point in $C$ by a red line. Take any $B_{i}, B_{j} \in B(1 \leqslant i < j \leqslant 2 n)$, they are connected to at most two points in $C$ by red lines. We can take a point $C_{k}$ in $C$ such that $B_{i} C_{k}$ and $B_{j} C_{k}$ are both blue lines, so $B_{i} B_{j}$ is a red line. Thus, the $2 n$ points in $B$ form a red complete $2 n$-sided polygon $K_{2 n}$, and together with $A_{0}$, they form a red complete $2 n+1$-sided polygon $K_{2 n+1}$. In this red complete $2 n+1$-sided polygon $K_{2 n+1}$, there is clearly a "windmill" with $n$ "blades", a contradiction!
From (i) and (ii), we get $f(n)=4 n+1(n \geqslant 2)$.
In summary, $f(n)=\left\{\begin{array}{ll}6, & n=1, \\ 4 n+1, & n \geqslant 2 .\end{array}\right.$