CS325 HW 7 Solved

Description

1. Let X and Y be two decision problems. Suppose we know that X reduces to Y in polynomial time. Which of the following can we infer? Explain.
   1. If Y is NP-complete then so is X.
   2. If X is NP-complete then so is Y.
   3. If Y is NP-complete and X is in NP then X is NP-complete.
   4. If X is NP-complete and Y is in NP then Y is NP-complete.
   5. If X is in P, then Y is in P.
   6. If Y is in P, then X is in P.
2. Consider the problem COMPOSITE: given an integer y, does y have any factors other than one and itself? For this exercise, you may assume that COMPOSITE is in NP, and you will be comparing it to the well-known NP-complete problem SUBSET-SUM: given a set S of n integers and an integer target t, is there a subset of S whose sum is exactly t?  Clearly explain whether or not each of the following statements follows from that fact that COMPOSITE is in NP and SUBSET-SUM is NP-complete:
   1. SUBSET-SUM ≤p
   2. If there is an O(n³) algorithm for SUBSET-SUM, then there is a polynomial time algorithm for COMPOSITE.
   3. If there is a polynomial algorithm for COMPOSITE, then P = NP.
   4. If P  NP, then no problem in NP can be solved in polynomial time.
3.  A Hamiltonian path in a graph is a simple path that visits every vertex exactly once. Prove that HAM-PATH = { (G, u, v ): there is a Hamiltonian path from u to v in G }  is NP-complete.  You may use the fact that HAM-CYCLE is NP-complete.
4. ) K-COLOR. Given a graph G = (V,E), a k-coloring is a function c: V -> {1, 2, … , k} such that c(u)  c(v) for every edge (u,v)   In other words the number 1, 2, .., k represent the k colors and adjacent vertices must have different colors.  The decision problem K-COLOR asks if a graph can be colored with at most K colors.
   1. The 2-COLOR decision problem is in P. Describe an efficient algorithm to determine if a graph has a 2-coloring.  What is the running time of your algorithm?

1. It is known that the 3-COLOR decision problem is NP-complete by using a reduction from SAT. Use the fact that 3-COLOR is NP-complete to prove that 4-COLOR is NP-complete.