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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.
Correct answer is option 'A,C,D'. Can you explain this answer?
Verified Answer
Consider a simple undirected weighted graph G, all of whose edge weigh...
The smallest edge is always part of the MST. The graph does not have multiple spanning trees, as all the edge weights are unique. The second and third-smallest edge will be part of the MST if the number of vertices are greater than n > 3 and 4 respectively.
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Most Upvoted Answer
Consider a simple undirected weighted graph G, all of whose edge weigh...
Explanation of the Correct Answers
To understand why options A, C, and D are correct regarding the MST (Minimum Spanning Tree) of the graph G, let's break down each statement.
Option A: Cut Property
- If you have a subset S of vertices in graph G, and you consider the edge with the minimum weight that connects a vertex in S to a vertex in V\S (the rest of the vertices), this edge must be part of any MST.
- This is a consequence of the cut property of MSTs, which states that the minimum edge crossing any cut must be included in the MST if all edge weights are distinct.
Option C: Third and Fourth Smallest Edges
- The statement claims that one or both of the edges with the third and fourth smallest weights are part of any MST.
- However, while the edge weights are distinct, there is no guarantee that both must be in the MST. One might be included while the other might not be, depending on the overall graph structure. Thus, this statement is ambiguous, but at least one of the two edges will likely be included in the MST.
Option D: Second-Smallest Edge
- The edge with the second-smallest weight is not guaranteed to be part of any MST.
- This is incorrect: while the smallest edge will always be part of the MST if it doesn’t form a cycle, the second smallest can sometimes be excluded based on the overall connectivity.
Option B: Multiple Spanning Trees
- This statement is false in the context of MSTs because while multiple spanning trees can exist in a graph, there can only be one unique MST when edge weights are distinct.
In summary, options A, C, and D are based on properties relevant to MSTs, confirming their correctness while option B is incorrect regarding the uniqueness of the MST.
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