For this problem, the answer is given for the pseudocode. But what would be the

Q: For this problem, the answer is given for the pseudocode. But what would be the

Solution: There are two loops in the given algorithm where a the while loop us the outer loop and the for loop is nested in it. T(n)= O(2^(n-2) * e) because in the worst case maximum possible path in a directed graph are 2^(n-2). here n is number of vertices and e is number of edges. I hope this helps if you find any problem. Please comment below. Don't forget to give a thumbs up if you liked it. :)

ID: 3706875 • Letter: F

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For this problem, the answer is given for the pseudocode. But what would be the proof and running time?

Problem 5. The max-bottleneck path problem: Given a weighted undirected graph G (V,E) with non- negative weights w(u,v) for each edge (u,v)E E, suppose the weight w(u,v) represents the max-capacity of a message sent along the edge (u,v). (Alternatively, you can consider the edges as roads and the weights as the heights of the bridges above these roads.) Thus, the bottleneck along a path (v_0,v_1,..v_k) is minisisk sw(vi-1,v¡)J.A max-bottleneck path between u and v is a path whose bottleneck is the largest among all paths connecting u and v. Given a starting vertex s, propose an algorithm in pseudocode to find the max-bottleneck path from s to every other node in V. Prove the correctness of your algorithm and analyze its running time Expert Answer nonymous answered this 377 answers Was this answer helpful? 0 1. Create a directed graph for the given undirected graph add directed edges(u->v, v->u) between every pair of vertices u & v 2. Initialize s as starting vertex 3. while there is a path from s to v: for each edge (u,v) remove u->v check v->u is there if no add v->u 4. Label each label whether or not reachable from s 5. Create a set B of edges which is the set of bottleneck edges if edge connects a reachable and a non reachable vertex Comment>

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Solution:

There are two loops in the given algorithm where a the while loop us the outer loop and the for loop is nested in it.

T(n)= O(2^(n-2) * e)

because in the worst case maximum possible path in a directed graph are 2^(n-2).

here n is number of vertices and e is number of edges.

I hope this helps if you find any problem. Please comment below. Don't forget to give a thumbs up if you liked it. :)

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For this problem, the answer is given for the pseudocode. But what would be the

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