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You are given a directed graph with (possibly negative) weighted edges, in which the shortest path between any two vertices is guaranteed to have at most edges. Give an algorithm that finds the shortest path between two vertices u and v in O(KE)time.

Short Answer

Expert verified

Bellman Ford algorithm in which the edges have negative weights and it finds the shortest path between two vertices inOKE time.

Step by step solution

01

 Step 1: Algorithm used for negative weighted directed graph

Bellman-Ford algorithmis an application of a single source shortest path, which is used forfinding the shortest distance from one vertex to other vertices of a weighted directed graph.

It is almost similar to Dijkstra's algorithm but Dijkstra's algorithm works only for the graph with a positive weight and Bellman-Ford algorithm works with graphs in which edges have negative weights in its graph.

02

Design the Algorithm

Bellman-Ford algorithm applies to the graph for finding the single source’s shortest path.A directed graph with positive and negative edge weight, and returns the length of the shortest cycle in the graph and the graph is acyclic, which takes time at mostOKE So, here the vertex A is the source vertex. Now take an array as a data structure to evaluate a single source’s shortest path between the source and the destination.

From A the distance of A is zero and take the distance of vertex A from each and every vertex is infinity. Now take A as the first vertex and evaluate the weight towards each vertex. Draw a directed positive and negative weighted graph as shown below:

Choose the next vertex from the vertices which have minimum weight and select that node as the second vertex. Then again evaluate the distance of it from every vertex and as get the minimum weight of the node and consider it as the main node. Through this the series of the vertex arises.

Here the vertex A is the source vertex. now take a minheap as a data structure for evaluate single source shortest path between the source and the destination.

From A the distance of A is zero and take the distance of vertex A from each and every vertex is infinity.

All vertices will be released many times in Bellman Ford algorithm.

Select every vertex one by one and put it into the array as a data structure one by one as shown in the figure.

Hence, the shortest distance from the vertex A to vertex D is evaluate in OKEtime.

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Most popular questions from this chapter

Shortest paths are not always unique: sometimes there are two or more different paths with the minimum possible length. Show how to solve the following problem in O((|V|+|E|)log|V|)time.

Input:An undirected graph G=(V,E);edge lengths le>0; starting vertex sV.

Output:A Boolean array for each node u , the entry usp[u]should be true if and only if there is a unique shortest path s to u (Note:usp[s]=true)

Professor F. Lake suggests the following algorithm for finding the shortest path from node to node t in a directed graph with some negative edges: add a large constant to each edge weight so that all the weights become positive, then run Dijkstra’s algorithm starting at node s , and return the shortest path found to node t .

Is this a valid method? Either prove that it works correctly, or give a counterexample.

Squares.Design and analyse an algorithm that takes as input an undirected graph G(V,E) and determines whether graph contains a simple cycle (that is, a cycle which doesn’t intersect itself) of length four. Its running time should be at mostO(V3) time.

You may assume that the input graph is represented either as an adjacency matrix or with adjacency lists, whichever makes your algorithm simpler.

Here's a proposal for how to find the length of the shortest cycle in an undirected graph with unit edge lengths. When a back edge, say (v,w), is encountered during a depth-first search, it forms a cycle with the tree edges from wtov. The length of the cyclelevel[v]-level[w+1] is where the level of a vertex is its distance in the DFS tree from the root vertex. This suggests the following algorithm:

• Do a depth-first search, keeping track of the level of each vertex.

• Each time a back edge is encountered, compute the cycle length and save it if it is smaller than the shortest one previously seen. Show that this strategy does not always work by providing a counterexample as well as a brief (one or two sentence) explanation.

In cases where there are several different shortest paths between two nodes (and edges have varying length),the most convenient of these paths is often the one with fewest edges. Forinstance, if nodes represent cities and edge lengths represent costs of flying between cities, theremight be many ways to get from cityto city t which all have the same cost. The mostconvenientof these alternatives is the one which involves the fewest stopovers. Accordingly, for a specific starting node S , define

bestu=minimum number of edges in a shortest path from S to u .

In the example below, thebestvalues for nodes S,A,B,C,D,E,Fare 0,1,1,1,2,2,3, respectively.

Give an efficient algorithm for the following problem.

Input:Graph G=V,E; positive edge lengths le; starting node sV.

Output: The values of bestu should be set for all nodesuV

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