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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  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 distin...
For 82a The answer should be Option A because edge 'e' is the lightest safe edge connecting X and Y so the minimum
spanning tree of G must contain 'e' (Greedy and optimal choice).
While B might seem correct but it is not always true. One such case is when G is not connected therefore there might not
be any path between 's' and 't'.
Since the question is about definitely true B is incorrect and A is the only correct option
Lets say AC =1 CD = 2 BD = 3 and AB=4
Then if s= A and t= B then AC is the lightest edge crossing X and Y where X = { A } and Y = { C, B, D}
But clearly AC is not on the shortest path from A to B. The shortest path is AB = 4.
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Most Upvoted Answer
Let S and t be two vertices in a undirected graph G=(V,E)having distin...
Explanation:

Minimum Weighted Spanning Tree:
- The edge e with the minimum weight amongst all edges connecting X and Y will definitely be a part of the minimum weighted spanning tree of the graph G.
- This is because in a minimum weighted spanning tree, the sum of the weights of the edges is minimized while still connecting all vertices in the graph.
- As e is the minimum weight edge connecting X and Y, it will be necessary for the spanning tree to connect these two partitions with the minimum possible weight.

Weighted Shortest Path from s to t:
- The edge e will also be a part of the weighted shortest path from s to t.
- This is because in a shortest path, we aim to minimize the total weight of the path from s to t.
- As e is the minimum weight edge connecting X and Y, it will be included in the shortest path to minimize the total weight from s to t.

Each Path from s to t:
- The edge e may not necessarily belong to each path from s to t.
- While it will be a part of the minimum weighted spanning tree and the weighted shortest path from s to t, it may not be present in all possible paths from s to t.
- Paths that do not include e may have a higher total weight compared to those that include e.

Weighted Longest Path from s to t:
- The edge e will definitely not be a part of the weighted longest path from s to t.
- In the longest path, we aim to maximize the total weight of the path from s to t.
- As e is the minimum weight edge connecting X and Y, it will not be included in the longest path as it minimizes the total weight.
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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 apartition of Vsuch that s ∈ X and t∈ Y. Consider the edge having the minimum weight amongst all those edges thathave one vertex in X and one vertex in YThe 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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