Let G be a weighted undirected graph and e be an edge with maximum weight in G. Suppose there is a minimum weight spanning tree in G containing the edge e. Which of the following statements is always TRUE?

**(A)** There exists a cutset in G having all edges of maximum weight.**(B)** There exists a cycle in G having all edges of maximum weight**(C)** Edge e cannot be contained in a cycle.**(D)** All edges in G have the same weight**Answer:** **(A)****Explanation:** **Background :** Given a connected and undirected graph, a spanning tree of that graph is a subgraph that is a tree and connects all the vertices together.

- A spanning tree has exactly V – 1 edges.
- A single graph can have many different spanning trees. A minimum spanning tree (MST) or minimum weight spanning tree for a weighted, connected and undirected graph is a spanning tree with weight less than or equal to the weight of every other spanning tree. The weight of a spanning tree is the sum of weights given to each edge of the spanning tree.
- There can be more that one possible spanning trees of a graph. For example, the graph in this question has 6 possible spanning trees.
- MST has lightest edge of every cutset. Remember Prim’s algorithm which basically picks the lightest edge from every cutset.

**Choices of this question :*** a) There exists a cutset in G having all edges of maximum weight* : This is correct. If there is a heaviest edge in MST, then there exist a cut with all edges with weight equal to heavies edge. See point 4 discussed in above background.

**b) There exists a cycle in G having all edges of maximum weight :** Not always TRUE. The cutset of heaviest edge may contain only one edge. In fact there may be overall one edge of heavies weight which is part of MST (Consider a graph with two components which are connected by only one edge and this edge is the heavies)

**c) Edge e cannot be contained in a cycle.** Not Always True. The curset may form cycle with other edges.

**d) All edges in G have the same weight** Not always True. As discussed in option b, there can be only one edge of heaviest weight.

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