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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...
Belongs to X and t belongs to Y. Let e be the lightest edge with one end in X and one end in Y . Let (X', Y') be the partition of V - {e} induced by the partition (X, Y) . Then, the edge e is a safe edge for the partition (X', Y').

Proof:
Let T be a minimum spanning tree of G. If e is not in T, then T is also a minimum spanning tree for G - {e}. Therefore, the claim holds in this case.

Now, let's assume that e is in T. Since e is the lightest edge with one end in X and one end in Y, there must be an edge f in T such that f is in X' and f is in Y'. We can see that e and f form a cycle in T, and since e is the lightest edge in this cycle, it must be the heaviest edge in the cycle. Therefore, by removing e from the cycle, we obtain a spanning tree that includes all vertices of G - {e}. Hence, (X', Y') is a valid partition induced by this spanning tree.

Therefore, e is a safe edge for the partition (X', Y'), and this completes the proof.
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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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