# A2 In an undirected weighted graph with distinct edge weights, both

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A2. In an undirected weighted graph with distinct edge
weights, both the lightest and the second lightest edge are
in some MST. Is this true or false? Justify your answer. [10
points]
NOTE: full explination is needed and answer in copiable
format
Solution
1.Because edge weights are distinct there is only a single
MST.
2.Let us consider as a1 and a2 be the lightest and second
lightest edge in Krushal\'s algorithm , a1 is always the first
edge added .since a2 cannot posible to create a cycle it
will necessarily to add next edge by krushal\'s algorithm
.By the correctness of krushal\'s algorithm ,the two lightest
edges are always in the MST

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A2. In an undirected weighted graph with distinct edge weights, both the lightest and the second lightest edge are in some MST. Is this true or false? Justify your answer. [10 points] NOTE: full explination is needed and answer in copiable format Solution Answer:- True 1.Because edge weights are d ...
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