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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?
Question Description
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? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about 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? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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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