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Let s and t be two vertices in a undirected graph G + (V, E) having distinct positive edge weights. Let [X, Y] be a partition of V such that s ∈ X and t ∈ Y. Consider the edge e having the minimum weight amongst all those edges that have one vertex in X and one vertex in Y The edge e must definitely belong to:
  • a)
    the minimum weighted spanning tree of G
  • b)
    the weighted shortest path from s to t
  • c)
    each path from s to t
  • d)
    the weighted longest path from s to t
Correct answer is option 'A'. Can you explain this answer?
Verified Answer
Let s and t be two vertices in a undirected graph G + (V, E) having di...
The minimum weight edge on any s-t cut is always part of MST. This is called Cut Property. This is the idea used in Prim's algorithm. The minimum weight cut edge is always a minimum spanning tree edge. Why B (the weighted shortest path from s to t) is not an answer? See below example, edge 4 (lightest in highlighted red cut from s to t) is not part of shortest path. 
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Most Upvoted Answer
Let s and t be two vertices in a undirected graph G + (V, E) having di...
Explanation:

To understand why the edge e must belong to the minimum weighted spanning tree of G, let's consider the other options and analyze their validity.

Option B: the weighted shortest path from s to t
The minimum weighted spanning tree is not necessarily the shortest path from s to t. It is possible for the shortest path to have edges that are not part of the minimum weighted spanning tree. Therefore, option B is not necessarily true.

Option C: each path from s to t
Similar to option B, the minimum weighted spanning tree is not necessarily part of every path from s to t. There can be multiple paths from s to t, and not all of them will include the minimum weighted spanning tree. Therefore, option C is not necessarily true.

Option D: the weighted longest path from s to t
The minimum weighted spanning tree is not designed to find the longest path between two vertices. Its purpose is to find the minimum weight connecting all vertices in the graph. Therefore, option D is not true.

Option A: the minimum weighted spanning tree of G
The minimum weighted spanning tree of G is defined as a tree that connects all vertices in G with the minimum total weight. It does this by selecting the edges with the minimum weight until all vertices are connected. By definition, the edge with the minimum weight that connects one vertex in X and one vertex in Y will be part of the minimum weighted spanning tree. Therefore, option A is true.

In conclusion, the edge e must belong to the minimum weighted spanning tree of G because it connects vertices from two different partitions and has the minimum weight among all such edges. This property makes it an essential part of the minimum weighted spanning tree.
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Let s and t be two vertices in a undirected graph G + (V, E) having distinct positive edge weights. Let [X, Y] be a partition of V such that s ∈ X and t ∈ Y. Consider the edge e having the minimum weight amongst all those edges that have one vertex in X and one vertex in Y The edge e must definitely belong to:a)the minimum weighted spanning tree of Gb)the weighted shortest path from s to tc)each path from s to td)the weighted longest path from s to tCorrect answer is option 'A'. Can you explain this answer?
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