Question

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4. a) Define the concept of NP-Completeness B) Show that there is a polynomial time algorithm that finds a longest path in a directed graph, under the condition that A is NP-complete and A has a polynomial time algorithm.

Answer 1

a. A decision problem L is NP-complete if:
1) L is in NP (Any given solution for NP-complete problems can be verified quickly, but there is no efficient known solution).
2) Every problem in NP is reducible to L in polynomial time (Reduction is defined below).

A problem is NP-Hard if it follows property 2 mentioned above, doesn’t need to follow property 1. Therefore, NP-Complete set is also a subset of NP-Hard set.

b.

The longest path problem for a general graph is not as easy as the shortest path problem because the longest path problem doesn’t have optimal substructure property. In fact, the Longest Path problem is NP-Hard for a general graph. However, the longest path problem has a linear time solution for directed acyclic graphs. The idea is similar to linear time solution for shortest path in a directed acyclic graph., we use Topological Sorting.

We initialize distances to all vertices as minus infinite and distance to source as 0, then we find a topological sorting of the graph. Topological Sorting of a graph represents a linear ordering of the graph Once we have topological order (or linear representation), we one by one process all vertices in topological order. For every vertex being processed, we update distances of its adjacent using distance of current vertex.

Following is complete algorithm for finding longest distances.
1) Initialize dist[] = {NINF, NINF, ….} and dist[s] = 0 where s is the source vertex. Here NINF means negative infinite.
2) Create a toplogical order of all vertices.
3) Do following for every vertex u in topological order.
………..Do following for every adjacent vertex v of u
………………if (dist[v] < dist[u] + weight(u, v))
………………………dist[v] = dist[u] + weight(u, v)