Draw a weighted graph with $5$ vertices has exactly two minimal spanning trees and justify that there are no other minimum spanning tree.
I could create a random graph which contain one minimal spanning tree but how to confirm it has another spanning tree?
Like the above graph has one minimal spanning tree. Another confusion is what does it mean when they say exactly two minimal spanning trees? Does those spanning tree should have same minimum cost? If not then isn't the latter spanning tree could be more or less cost?
Let $G = (V, E, w)$ be a weighted graph with $V = \{v_1, v_2, v_3, v_4, v_5\}$, $E = \{\{v_1, v_3\}, \{v_4, v_3\}, \{v_5, v_3\}, \{v_2, v_3\}, \{v_1, v_4\}\}$ and $$\begin{align} w(\{v_1, v_3\}) &= 1 \\ w(\{v_4, v_3\}) &= 2 \\ w(\{v_5, v_3\}) &= 3 \\ w(\{v_2, v_3\}) &= 4 \\ w(\{v_1, v_4\}) &= 2 \end{align}$$
Then, it is easy to see $G$ has $3$ spanning trees by removing one of the $3$ edges $\{v_1, v_3\}, \{v_4, v_3\}, \{v_1, v_4\}$. Note that they have total weight $11, 10, 10$ respectively. Therefore, $G$ has exactly $2$ minimum spanning trees.