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

Time complexity of Ford fulkerson Algorithm is2 pointsO (E * |f|) , where f is the maximum flow and E is no of edges in network.O (E + |f|) , where f is the maximum flow and E is no of edges in network.O (E / |f|) , where f is the maximum flow and E is no of edges in network.O (E - |f|) , where f is the maximum flow and E is no of edges in network.

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.

Consider the Bellman-Ford algorithm with FIFO queue. What is the running time of the whole Bellman-Ford algorithm, if the queue is (correctly but inefficiently) implemented in such a way that the running time of one DEQUEUE operation and the running time of one ENQUEUE operation are both O(n)O(n) time? Select the most accurate bound.Question 4Select one:O(mn)O(mn)O(mnlogn)O(mnlog⁡n)O(n3)O(n3)O(mn2)O(mn2)O(m2n)

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

Time complexity of Ford fulkerson Algorithm is