II. (50 points) In a convex $n$-sided rose garden ($n \geqslant 4$), there is 1 red rose planted at each of the $n$ vertices. There is a straight path between every two red roses, and these paths do not have the situation of "three lines intersecting at one point". They divide the garden into many non-overlapping regions (triangles, quadrilaterals, etc.), and each region has one white rose (or black rose) planted in it. Try to find the expression for the total number of roses $f(n)$ in the rose garden.

 untranslated part:
These straight paths do not have the situation of "three lines intersecting at one point". They divide the garden into many non-overlapping regions (triangles, quadrilaterals, etc.), and each region has one white rose (or black rose) planted in it. Try to find the expression for the total number of roses $f(n)$ in the rose garden. 

(Note: The untranslated part is a repetition of the already translated content, so it is not necessary to translate it again.)
II. Solution: The straight paths in the rose garden form a convex $n$-sided polygon and its diagonals $\left(k_{n}\right)$. It is easy to know that the number of diagonals in the graph is $C_{n}^{2}-n$. Furthermore, given that no three diagonals intersect at a single point, the number of intersection points of the diagonals inside the polygon is $C_{n}^{4}$. We first find the number of regions $g(n)$ inside the $n$-sided polygon, and then we have $f(n)=g(n)+n$.

Now, remove diagonal $l_{1}$, and let $l_{1}$ intersect with other diagonals at $d_{1}$ points. The graph $K_{n}$ is reduced by $d_{1}+1$ regions. Denote the remaining graph as $G_{1}$.

In $G_{1}$, remove diagonal $l_{2}$, and let $l_{2}$ intersect with other diagonals (excluding $l_{1}$) at $d_{2}$ points in $G_{1}$. The graph $G_{1}$ is reduced by $d_{2}+1$ regions. Denote the remaining graph as $G_{2}$.

Proceeding in this manner, each time remove one diagonal from the remaining graph. When the last diagonal is removed, it no longer intersects with any other remaining diagonals, and the graph is reduced by 1 region. In the end, only 1 region remains.
$$
g(n)=\sum_{m=1}^{C_{n}^{2}-n}\left(d_{m}+1\right)+1=\sum_{m=1}^{C_{n}^{2}-n} d_{m}+\left(C_{n}^{2}-n\right)+1=C_{n}^{4}+C_{n}^{2}-n+1=C_{n}^{4}+C_{n-1}^{2} .
$$