Consider the following network. With the indicated link costs, use Djikstra's shortest- path algorithm to compute the table of shortest paths from A to all other network nodes. Show how the algorithm works by filling out the table below. Use column “N" for “all visited nodes in current step", and each row for "distance and predecessor of each destination node once a new node is visited".

I have spent hours on internet to find an example of iterative lengthening search algorithm but found nothing. I need it ASAP. Can anyone help me? Please. It is its definition: Iterative lengthening search is an iterative analogue of uniform-cost search. The basic idea is to use increasing limits on path cost. If a node is generated whose path cost exceeds the current limit, it is immediately discarded. For each new iteration, the limit is set to the lowest path cost of any node discarded in the previous iteration. I dont understand many things in this definition. 1-What is the first limit before we start? 2-Do we delete the discarded node from the graph or what? 3-What is the generated node? Is it expanded or is it just found as neigbor? 4-What is "lowest path cost of any node discarded"?? Path from where to where?? From start to that node or what? 5-What does path cost mean in this problem? Is it from source to the new node or between current node and next chosen node? I am literally…

Subject : Algorithm and Data Structure

5.04-4. Bellman Ford Algorithm - a change in DV (1, part 4). Consider the network below, and suppose that at t=0, the link between nodes b and c goes down. And so at t=0, node b recomputes its distance vector (DV) and sends out its new DV (as needed). At t=1 this new DV is received at b's neighbors, who then perform their calculation and send out their new DVs (as needed); these new DVs arrive at their neighbors at t=2, and so on. What is the last time in this network at which a DV calculation will take place as a result of the link change at t=0? U O O o 1 1 3 2 at t=0 the link (with a cost of 1) between nodes b and c goes down A 8 1 6 compute 1 1 1 1 an essentially infinite amount of time; this is the count-to-infinity problem 1 Second