The seventh question: On a plane, there is a complete graph of order $\mathrm{n}(\mathrm{n} \geq 3)$. The edges of the graph are colored with three colors. Now assume that in the complete graph, we can always select $\mathrm{k}$ edges, and change the colors of these $\mathrm{k}$ edges to one of the given three colors, such that the graph can also be connected by this color. If the goal can always be achieved regardless of the initial coloring, find the minimum value of $\mathrm{k}$.

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The seventh question: On a plane, there is a complete graph of order $\mathrm{n}(\mathrm{n} \geq 3)$. The edges of the graph are colored with three colors. Now assume that in the complete graph, we can always select $\mathrm{k}$ edges, and change the colors of these $\mathrm{k}$ edges to one of the given three colors, such that the graph can also be connected by this color. If the goal can always be achieved regardless of the initial coloring, find the minimum value of $\mathrm{k}$.
Question 7:
Solution: First, prove that $k \geq \left[\frac{n}{3}\right]$.
Assume the three colors are $1, 2, 3$, and the $n$ vertices of the $n$-order complete graph are divided into three sets $A, B, C$, with $|A| = |B| = \left[\frac{n}{3}\right]$.

Perform the following coloring: the edges between the points in set $A$ are all colored 1; the edges between the points in set $B$ are all colored 2; the edges between the points in set $C$ are all colored 3; the edges between the points in set $A$ and $B$ are all colored 2; the edges between the points in set $B$ and $C$ are all colored 3; the edges between the points in set $C$ and $A$ are all colored 1.

Thus, if the final connected graph formed by the edges colored 1, since the edges from the points in set $B$ are all colored 2 or 3, the number of recolored edges is no less than $|B| = \left[\frac{n}{3}\right]$.

Similarly, if the final connected graph formed by the edges colored 2 or 3, the number of recolored edges is no less than $|A|$ (or $|C|$) $\geq \left[\frac{n}{3}\right]$.
Therefore, in the above scenario, it must be that $k \geq \left[\frac{n}{3}\right]$.
Next, prove that $k = \left[\frac{n}{3}\right]$ satisfies the condition.
Use mathematical induction on $n$.
When $n = 3$, the conclusion is obviously true.
Assume the conclusion holds for $n-1$ ($n \geq 4$). Now consider the case with $n$ points.
If there exists a point such that all its outgoing edges are the same color, then the complete graph is connected by that color. The conclusion is obviously true. Now assume that each point has edges of at least two different colors.
(1) If there is a point in the complete graph whose outgoing edges have three different colors, according to the induction hypothesis, it is possible to change $\left[\frac{n-1}{3}\right] \leq \left[\frac{n}{3}\right]$ edges to make the graph connected by one color. The conclusion holds.
(2) If all points in the complete graph have exactly two different colors for their outgoing edges, let $A$ be the set of points whose outgoing edges are all colored 1 or 2; let $B$ be the set of points whose outgoing edges are all colored 2 or 3; let $C$ be the set of points whose outgoing edges are all colored 3 or 1.
Without loss of generality, assume $|A| \leq |B| \leq |C|$. Then $|A| \leq \left[\frac{n}{3}\right]$.
(1) If $B = \emptyset$, then $A = \emptyset$, so all edges in the complete graph are colored 1 or 3. Choose any one of these points, which has edges of color 1 and color 3. According to (1), the remaining points can be connected by changing $\left[\frac{n-1}{3}\right]$ edges to color 1 or 3, making the entire complete graph connected by that color.
(2) If $B \neq \emptyset$, then the edges from points in set $B$ to points in set $C$ are all colored 3. Hence, $B \cup C$ is connected by color 3. At this point, choose any point in set $B$, and change all the edges from points in set $A$ to this point to color 3, making the entire complete graph connected by color 3. Since $|A| \leq \left[\frac{n}{3}\right]$, the conclusion holds.
By mathematical induction, the conclusion holds.
In summary, the minimum value of $k$ that satisfies the condition is $\left[\frac{n}{3}\right]$.
Note: If the $n$-order complete graph is two-colored, then the complete graph must be connected by one color.
In fact, use induction on $n$. When $n = 3$, the conclusion is obviously true.
Assume the conclusion holds for $n-1$. Now consider the case with $n$ points.
If there exists a point whose outgoing edges are all the same color, then the complete graph is connected by that color; if a point has edges of two different colors, according to the induction hypothesis, the remaining $n-1$ points must be connected by one color, making the entire $n$-order complete graph connected by that color.
By mathematical induction, the conclusion holds.