G t f 2 points in a weighted undirected tree g v e

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Unformatted text preview: and breadth-first search finds it. (g) T F [2 points] In a weighted undirected tree G = (V, E, w), depth-first search from a vertex s finds single-source shortest paths from s (via parent pointers) in O(V + E ) time. Solution: True. In a tree, there is only one path between two vertices, and depth-first search finds it. (h) T F [2 points] If a graph represents tasks and their interdependencies (i.e., an edge (u, v ) indicates that u must happen before v happens), then the breadth-first search order of vertices is a valid order in which to tackle the tasks. Solution: No, you’d prefer depth-first search, which can easily be used to produce a topological sort of the graph, which would correspond to a valid task order. BFS can produce incorrect results. (i) T F [2 points] Dijkstra’s shortest-path algorithm may relax an edge more than once in a graph with a cycle. Solution: False. Dijkstra’s algorithm always visits each node at most once; this is why it produces an incorrect result in the presence of negative-weight edges. (j) T F [2 points] Given a weighted directed graph G = (V, E, w) and a source s ∈ V , if G has a negative-weight cycle somewhere, then the Bellman-Ford algorithm...
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