Problem 3. Counting Paths (22 points) Let G be a directed acyclic graph with n v

Question

Problem 3. Counting Paths (22 points) Let G be a directed acyclic graph with n vertices, and let s,t be two labeled vertices in G ( . (a) (6 points) Argue that there can be as many as 2-2 directed paths from s to f (b) (16 points) Give an O(n +m) time algorithm to count the number of directed paths which start at s and end at t, in any directed acyclic graph with n vertices and m edges. In addition to a clear description of the algorithm, make sure to include explanations for correctness and running time.

Explanation / Answer

Solution:

Suppose we have a directed graph G(V, E) in which there is no cycle.

so now,

suppose the graph has n number of vertices

then

the number of vertices in the paths of s to t will be n-2

since s ant t are the source and the destination

and the choices at each node here is binary

This means that the number of paths will be of power of 2

and n-2 vertices are there in between

This means the maximum number of paths= 2^(n-2)

So number of paths<= 2^(n-2)

b)

Algorithm:

Correctness:

In the above algorithm we are visiting every vertex can counting the number of paths that particular vertex provides, and for all the verticess it is being added. That is why the result will be a coorect output:

running time:

The running time of the above algorithm will be O(n+m), where n is the number of vertices in the graph and m is the number of edges.

Since we are visiting each vertex and every number of edges is considered for the path that is why O(n+m)

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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