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Question: Prove that for the array prev computed by Dijkstra's algorithm, the edges {u,prepu}(forall∈v)form a tree.

Short Answer

Expert verified

All the edges {u,prepu}(foralll^v)form a tree follows this condition if the nodes are connected so there is no cycle.

Step by step solution

01

Step-1: Dijkstra’s Algorithm.

Dijkstra’s algorithm is a single-source shortest path problem. It is used in both directed and undirected graph with non negative weight.

Dijkstra's method to find shortest path:

  • Begin the initialization process at the root node.
  • Update the cost of the adjacent nodes in the table by identifying the adjacent nodes.
  • From the table, find the node with the lowest cost and repeat step-2 until all nodes have been traversed.
02

Step-2: Explanation of Dijkstra’s Algorithm.q p

To prove that a graph is a tree, we must show:

1. All of the nodes are linked together.

The definition of Dijkstra's algorithm demonstrates this. If there are some node pq , then there is a path from q to p if prev[p] Equals q. If prev[m] for some node m q , then there is a route from q to m. As a result, if there is an edge prev[r], for each node r, there is always apath from q to r. As a result, node q is connected to all other nodes.

2. There is not any cycle

Contradiction can be used to establish this. Assume that root q,i.e prev[q] , does not have a parent. It is also know that prev[r] is the alone parent of all other nodes r. If node v1,v2,...,vnhas a cycle, let's assume v1=prevv2,v2=prev[v3],...,vn=prevv1. None of nodes are linked to q, which violates the previousstatement. As a result, there is no cycle.

Hence, the edges {u,prepu}(foralll^v)form a tree if all the nodes are connected and there is no cycle.

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Most popular questions from this chapter

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 s∈V.

Output: The values of bestu should be set for all nodesu∈V

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.

Consider a directed graph in which the only negative edges are those that leaves; all other edges are positive. Can Dijkstra's algorithm, started at s, fail on such a graph? Prove your answer.

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 s∈V.

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)

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