7.18 Given 7 points in a plane, how many line segments must be connected between them at least, so that any 3 points have a line segment between two of them? Try to provide a graph that meets the requirement.
[Solution 1] The right figure shows 9 line segments connecting 7 points and satisfying the requirements of the problem, hence the minimum value we seek is no greater than 9.

To prove that in any graph satisfying the requirements, there must be at least 9 line segments:
(1) If there exists a point $A$ from which at most 1 line segment is drawn, then among the 5 points not connected to $A$, every pair of points must be connected, resulting in 10 line segments.
(2) If each point has at least 2 line segments and point $A$ has exactly 2 line segments $A B, A C$, then the 4 points not connected to $A$ must have 6 line segments among them. Point $B$ must have at least 1 additional line segment, so there are at least 9 line segments in total.
(3) If each point has at least 3 line segments, then the 7 points together have at least 21 line segments. Since each line segment is counted twice, the graph must have at least 11 line segments.
In summary, at least 9 line segments are required.
[Solution 2] We only need to prove that any graph satisfying the requirements must have at least 9 line segments.
Assume there is no line between points $A$ and $B$, then each of the points $C, D, E, F, G$ must be connected to at least one of $A$ or $B$, resulting in at least 5 line segments. The 5 points can form 10 different groups of three points, and each group of three points must have at least 1 line segment, resulting in at least 10 line segments. In this counting process, each line segment is counted exactly 3 times, so there must be at least 4 different line segments among the 5 points. Therefore, the graph must have at least 9 line segments.
[Solution 3] Let point $A$ be the point among the 7 points that has the fewest line segments, with a total of $k$ line segments $A B_{j}$, $j=1,2, \cdots, k$. Then the remaining $6-k$ points are not connected to $A$, so every pair of these points must be connected. Thus, the total number of line segments in the graph
$$
\begin{aligned}
S & \geqslant \frac{1}{2}\{k(k+1)+(6-k)(5-k)\}=\frac{1}{2}\left(2 k^{2}-10 k+30\right) \\
& =k^{2}-5 k+15=\left(k-\frac{5}{2}\right)^{2}+\frac{35}{4} \geqslant 8 \frac{3}{4},
\end{aligned}
$$

which means there must be at least 9 line segments.
On the other hand, the graph given in Solution 1 shows that 9 line segments are achievable, so the minimum number of line segments required is 9.