# Week 2 Problem Set #2 Quiz Answer

In this article i am gone to share Coursera Course Shortest Paths Revisited, NP-Complete Problems and What To Do About Them Week 2 Problem Set #2 Quiz Answer with you..

## Shortest Paths Revisited, NP-Complete Problems and

What To Do About Them Week 2 Quiz Answer

**Also visit:** Week 1 Problem Set #1 Quiz Answer

## Week 2 Problem Set #2 Quiz Answer

Question 1)

Which of the following statements cannot be true, given the current state of knowledge?

- Some NP-complete problems are polynomial-time solvable, and some NP-complete problems are not polynomial-time solvable.
- There is an NP-complete problem that is polynomial-time solvable.
- There is an NP-complete problem that can be solved in O(nlogn) time, where n is the size of the input.
- There is no NP-complete problem that can be solved in O(nlogn) time, where n is the size of the input.

Question 2)

Let TSP1 denote the following problem: given a TSP instance in which all edge costs are positive integers, compute the value of an optimal TSP tour. Let TSP2 denote: given a TSP instance in which all edge costs are positive integers, and a positive integer T, decide whether or not there is a TSP tour with total length at most T. Let HAM1 denote: given an undirected graph, either return the edges of a Hamiltonian cycle (a cycle that visits every vertex exactly once), or correctly decide that the graph has no such cycle. Let HAM2 denote: given an undirected graph, decide whether or not the graph contains at least one Hamiltonian cycle.

- If TSP2 is polynomial-time solvable, then so is TSP1. If HAM2 is polynomial-time solvable, then so is HAM1.
- Polynomial-time solvability of TSP2 does not necessarily imply polynomial-time solvability of TSP1. Polynomial-time solvability of HAM2 does not necessarily imply polynomial-time solvability of HAM1.
- If TSP2 is polynomial-time solvable, then so is TSP1. But, polynomial-time solvability of HAM2 does not necessarily imply polynomial-time solvability of HAM1.
- Polynomial-time solvability of TSP2 does not necessarily imply polynomial-time solvability of TSP1. But, if HAM2 is polynomial-time solvable, then so is HAM1.

Question 3)

Assume that PENP. Consider undirected graphs with nonnegative edge lengths. Which of the following problems can be solved in polynomial time?

Hint: The Hamiltonian path problem is given an undirected graph with n vertices, decide whether or not there is a (cyclefree) path with n-1 edges that visits every vertex exactly once. You can use the fact that the Hamiltonian path problem is NP-complete. There are relatively simple reductions from the Hamiltonian path problem to 3 of the 4 problems below.

- For a given source s and destination t, compute the length of a shortest s-t path that has exactly n – 1 edges (+∞ or too, if no such path exists). The path is allowed to contain cycles.
- Amongst all spanning trees of the graph, compute one with the smallest possible number of leaves.
- Amongst all spanning trees of the graph, compute one with the minimum-possible maximum degree. (Recall the degree of a vertex is the number of incident edges.)
- For a given source s and destination t, compute the length of a shortest s-t path that has exactly n – 1 or(+∞ edges or too, if no such path exists). The path is not allowed to contain cycles.

Question 4)

Choose the strongest true statement.

- If the minimum-size vertex cover problem can be solved in time O(Tn)) in bipartite graphs, then the maximumsize independent set problem can be solved in time O(T(n)) in bipartite graphs.
- If the minimum-size vertex cover problem can be solved in time O(T(n) in general graphs, then the maximumsize independent set problem can be solved in time O(T(n) in general graphs.
- If the maximum-size independent set problem can be solved in time (T(n) in general graphs, then the minimum-size vertex cover problem can be solved in time O(T(n)) in general graphs.
- All three of the other assertions are true.

Question 5)

Which of the following statements is true?

- Consider a TSP instance in which every edge cost is the Euclidean distance between two points in the place (just like in Programming Assignment #5). Deleting a vertex and all of its incident edges cannot increase the cost of the optimal (i.e., minimum sum of edge lengths) tour.
- Consider a TSP instance in which every edge cost is negative. The dynamic programming algorithm covered in the video lectures might not correctly compute the optimal (i.e., minimum sum of edge lengths) tour of this instance.
- Consider a TSP instance in which every edge cost is negative. Deleting a vertex and all of its incident edges cannot increase the cost of the optimal (i.e., minimum sum of edge lengths) tour.
- Consider a TSP instance in which every edge cost is either 1 or 2. Then an optimal tour can be computed in polynomial time.