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)?

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?

Consider the Bellman-Ford algorithm with FIFO queue and view its computation in the following stages:Stage 1: The source vertex ss is taken from the queue and Relax operation is applied to each edge outgoing from ss. The neighbours of ss have their shortest-path estimates set to finite values and are added to the queue.Stage 2: The neighbours of ss (added to the queue in Stage 1) are taken from the queue one by one and Relax operation is applied to each edge outgoing from those vertices. (Some vertices might be added to the queue during this computation.) A general stage i≥2i≥2: The vertices added to the queue in Stage i−1i−1 are taken from the queue one by one and Relax operation is applied to each edge outgoing from those vertices. (Some vertices might be added to the queue during this computation.)Select all correct statements, assuming that there is no negative cycle reachable from the source vertex. (Negative marks for selecting incorrect statements.)Hint: Refer to the analysis of the basic Bellman-Ford algorithm (slides 25-26) and observe that the stages of  Bellman-Ford-with-FIFO (as defined above) may be viewed as corresponding to the rounds (the iterations of the outer loop) of the  basic Bellman-Ford.Question 3Select one or more:There are at most nn stages during the whole computation.There may be more than nn stages during the whole computationThe source vertex may be added again to the queue after it is removed at the beginning of the computation.During one stage of the computation, each vertex is added to the queue at most once.During one stage of the computation, there may be two or more Relax operations applied to the same edge.The number of Relax operations during one stage of the computation is O(n)O(n).If a shortest path from the source vertex ss to a vertex vv has 33 edges, then the shortest-path estimate for vv is set to the correct shortest-path weight from ss to vv by the end of the 3-rd stage.  During one stage of the computation, for each vertex vv, the shortest-path estimate d[v]d[v] at this vertex can be updated more than once.If there are two shortest paths from the source vertex ss to a vertex vv, and one of them has 44 edges while the other has 55 edges, then Bellman-Ford-with-FIFO can compute either of them.

What is the key concept used in Bellman-Ford Algorithm to ensure the optimality of the solution?

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

What is the running time of Bellman Ford Algorithm when graph is Complete graph*1 pointO(V2)O(O(V3))O(VE)O(V)