Which of the following are true about the Bellman-Ford single source shortest path algorithm? Select all that apply.If the input graph has 𝑛n vertices and 𝑚m edges, the algorithm will always terminate in time 𝑂(𝑛𝑚)O(nm).The algorithm is only guaranteed to be correct when the edge weights are non-negative.The algorithm can detect if a graph has a negative weight cycle.If in a round of Bellman-Ford all edges are relaxed and no distance value is updated, then no distance value will be updated for the rest of the algorithm.Submit
Question
Which of the following are true about the Bellman-Ford single source shortest path algorithm? Select all that apply.If the input graph has 𝑛n vertices and 𝑚m edges, the algorithm will always terminate in time 𝑂(𝑛𝑚)O(nm).The algorithm is only guaranteed to be correct when the edge weights are non-negative.The algorithm can detect if a graph has a negative weight cycle.If in a round of Bellman-Ford all edges are relaxed and no distance value is updated, then no distance value will be updated for the rest of the algorithm.Submit
Solution
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True. The Bellman-Ford algorithm's time complexity is indeed O(nm), where n is the number of vertices and m is the number of edges. This is because, in the worst case, the algorithm relaxes all edges of the graph for every vertex.
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False. The Bellman-Ford algorithm can handle graphs with negative edge weights. The only requirement is that there are no negative weight cycles, as the algorithm cannot produce a correct result in such cases.
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True. One of the key features of the Bellman-Ford algorithm is its ability to detect negative weight cycles. If the algorithm is run for n iterations (where n is the number of vertices) and it is still able to relax an edge, it means that there is a negative weight cycle.
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True. If in a round of Bellman-Ford all edges are relaxed and no distance value is updated, then no distance value will be updated for the rest of the algorithm. This is because the algorithm has already found the shortest paths to all vertices.
Similar Questions
Shortest Path Problem Analysis: • Compare and contrast Dijkstra's algorithm with the Bellman-Ford algorithm in terms of their applicability, time complexity, and handling of negative edge weights. • Provide a scenario where Dijkstra's algorithm fails but the Bellman-Ford algorithm succeeds in finding the shortest path.
If a graph contains a negative cycle, what does Bellman-Ford Algorithm detect?
Let G be a directed acyclic graph. If we have the topological ordering of G, we can modify the Bellman's Ford algorithm to be a one-pass Bellman's Ford algorithm. It works as follow:For each vertex u of G in topological order, we relax the cost of u's outgoing edges. for each vertex u in topological order for each edge e // e = (u -> v) relax(e)Since the shortest path to vertex u is already found, we can subsequently update the shortest path of its adjacent vertices when we follow the topological order.What is the time complexity for the algorithm above?
Let G be a tree. Label the edges in BFS order, e.g., the first edge explored is given the label 1, the second edge explored is given the label 2, etc. As in the previous question, we will try to modify the Bellman's Ford algorithm to be a one-pass Bellman's Ford algorithm as follows:For each vertex u of G in BFS-order, we relax the cost of u's outgoing edges. for each vertex u in BFS order for each edge e // e = (u -> v) relax(e)What is the running time of this combined algorithm (i.e., finding the BFS ordering and running this Bellman-Ford variant)?
What is the running time of Bellman Ford Algorithm when graph is Complete graph*1 pointO(V2)O(O(V3))O(VE)O(V)
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