If image is too small please zoom in. Let u, v be two vertices in the directed g

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Let u, v be two vertices in the directed graph G. After running depth-first search (DFS) algorithm on G, it is possible to get the following discovery and finishing times: u.d = 4, u.f = 10, v.d = 8, and v.f = 12. BFS(G, s) computes shortest-paths from s to each vertex v reachable from s. If the graph G is sparse, then we can represent G using a binary search tree. Let G(V, E) be a directed graph such that a, b, c elementof G.V and (a, b), (b, c), (c, a) elementof. Then a, b, c must belong to the same strongly connected component. The height of a binary search tree with 15 nodes is greater than or equal to 3 and less than or equal to 14. A Minimum Spanning Tree is an acyclic subset of edges T subsetorequalto G.E, where all the edges in T are connected and the weight of the tree is minimized.

Explanation / Answer

The true statements are

If the Graph G is sparse, then we can rerepresent G using a binary tree.

The height of binary search tree with 15 nodes is greater than or equal to 3 and less than or equal to 14.

BFS(G,s) computes shortest-paths from s to each vertex v reachable from s.

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