Question

Consider a simple undirected weighted graph G, all of whose edge weights are distinct. Which of the following statements about the MST of G is/are true?

a)Suppose S⊆V be such that S ≠ θ and S ≠ V. Consider the edge with min weight such that one of its vertices is in S and the other in V\S. Such an edge will always be part of any MST of G.
b)G can have multiple spanning trees.
c)One or both the edges with the third smallest and the fourth-smallest edges are part of any MST of G.
d)The edge with the second-smallest weight is always part of any MST of G.

Answer 1

In a simple undirected weighted graph, the MST has certain properties. The edge with the minimum weight connecting a vertex in S and a vertex in V\S is always part of the MST.

Step-by-step explanation:

a) The statement is true. In the Minimum Spanning Tree (MST) of a graph, the edge with the minimum weight that connects a vertex in S and a vertex in V\S must be included. This is because an MST should have a path that connects all vertices in the graph, and including this edge is the most efficient way to do so.

b) The statement is true. A graph can have multiple spanning trees if there are different ways to connect all the vertices with the minimum total weight of edges. This happens when there are multiple edges with the same minimum weight.

c) The statement is false. The MST may or may not include the edges with the third smallest and fourth-smallest weights. The MST includes the edges with the minimum weight until all vertices are connected, but does not necessarily include the edges with the next few smallest weights.

d) The statement is true. In any MST, the edge with the second-smallest weight is guaranteed to be included. This is because after including the minimum-weight edge, the next lightest edge will always be selected to form the MST, and so on.