(6 pts) 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? Give a YES/NO answer.
If Y is NP-complete then so is X.
If X is NP-complete then so is Y.
If Y is NP-complete and X is in NP then X is NP-complete.
If X is NP-complete and Y is in NP then Y is NP-complete.
If X is in P, then Y is in P.
If Y is in P, then X is in P.
(4 pts) 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:
SUBSET-SUM ≤p COMPOSITE.
If there is an O(n3) algorithm for SUBSET-SUM, then there is a polynomial time algorithm for
COMPOSITE.
If there is a polynomial algorithm for COMPOSITE, then P = NP.
If P NP, then no problem in NP can be solved in polynomial time.
(8 pts) 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.
(12 pts) 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) E. 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.
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?
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.
