Question

Consider a simple undirected weighted graph \(G\), all of whose edge weights are distinct. Which of the following statements about the minimum spanning trees of \(G\) is/are TRUE?

• A. The edge with the second smallest weight is always part of any minimum spanning tree of \(G\).
• B. One or both of the edges with the third smallest and the fourth smallest weights are part of any minimum spanning tree of \(G\).
• C. Suppose \( S \subseteq V \) be such that \( S \neq \emptyset \) and \( S \neq V \). Consider the edge with the minimum weight such that one of its vertices is in \( S \) and the other in \( V \setminus S \). Such an edge will always be part of any minimum spanning tree of \( G \).
• D. \( G \) can have multiple minimum spanning trees.

Hint:

When working with minimum spanning trees, remember that the cut property guarantees the inclusion of the minimum weight edge crossing any cut. Distinct edge weights ensure a unique MST.

Answer 1

The Correct Option is A, B, C

Let's analyze each statement:

Option (A) - True:
- The second smallest edge is always part of any minimum spanning tree (MST) if it does not form a cycle. This is a consequence of Kruskal's algorithm: the second smallest edge is always added to the MST if it doesn't form a cycle, since adding it will reduce the total weight of the tree without violating the minimum weight requirement.

Option (B) - True:
- The third and fourth smallest edges might or might not be part of the MST. It is possible that one or both of these edges are included in the MST depending on their positioning and whether they create a cycle. However, it's also possible for neither edge to be included. The presence of these edges in the MST depends on the structure of the graph, but at least one of them can be part of the MST if it does not create a cycle.

Option (C) - True:
- This statement is true because the edge with the minimum weight between two subsets \( S \) and \( V \setminus S \) is guaranteed to be part of the MST. This is a direct application of the cut property of MSTs, which states that for any cut in the graph, the minimum weight edge crossing the cut must be part of the MST.

Option (D) - False:
- In a graph with distinct edge weights, the MST is unique. This is a consequence of Kruskal's and Prim's algorithms. If all edge weights are distinct, there is no ambiguity in choosing the minimum weight edges for the spanning tree, and thus the MST is unique.

Thus, the correct answers are (A), (B), and (C).