Time complexity of Hamiltonian circuit is O( n22nn22n) where n is number of vertices. Hence, NPH.

This is a statement not an explanation. Please explain the reason for it.

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- Is NPC and hence NP hard.
- Is again NP hard (optimization version is NP hard and decision version is NPC). Ref: http://stackoverflow.com/questions/3907545/how-to-understand-the-knapsack-problem-is-np-complete
- Is in P. See the algorithm here based on DFS: http://en.wikipedia.org/wiki/Biconnected_component
- NPC and hence NP hard.

Correct Answer: $C$

Assuming that you know all the terminologies of the question, here are the answers:

**Let V are the number of VERTICES and E are number of EDGES.**

A: Time complexity of Hamiltonian circuit is O( $n^{2}2^{n}$) where n is number of vertices. Hence, NPH.

The Hamilton circuit problem is polynomial time reducible to 3-SAT.

Now, since 3-SAT is NP complete therefore, Hamilton circuit problem also become np complete. Means NP hard.

B: Time complexity of 0/1 Knapsack is O(nW) where W is size of knapsack.

While the decision problem is NP-complete, the optimization problem is NP-hard, its resolution is at least as difficult as the decision problem, and there is no known polynomial algorithm which can tell, given a solution, whether it is optimal (which would mean that there is no solution with a larger *V*, thus solving the NP-complete decision problem).

Source: https://en.wikipedia.org/wiki/Knapsack_problem

D: Time complexity of Graph coloring is O$(V^{2}+E)$

This is NP-Complete. Reason : https://math.stackexchange.com/questions/125136/how-is-the-graph-coloring-problem-np-complete

C: Time Complexity of finding bi-connected components of a graph is O$(V+E)$. It is straightforward P time taking complexity.

**Hence C is the answer.**

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