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GATE | GATE-CS-2005 | Question 83

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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



Answer: (A)

Explanation:

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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Last Updated : 28 Jun, 2021
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