CS302 Lecture notes -- NP Completeness

This is not a complete treatment of NP-Completeness. Like the Halting Problem lecture notes, they introduce you to a concept that you will see later in your CS careers and will provide you with fodder for endless conversations around the family dinner table.

There is a very good introduction to complexity theory and P vs NP in wikipedia's notes. I suggest you take a read. In particular, their Venn diagram of P, NP, NP-Hard and NP-Complete is very nice. Learn it.

P, NP, NP-Complete and NP-Hard are sets of problems, defined as follows:

In this lecture, we are going to see what it takes to prove that problems belong to these sets. Suppose you have a problem to solve, and you want to know its complexity class. This takes two steps:

  1. Prove that it is in NP. Typically the problem is couched as a yes or no problem involving a data structure, such as ``does there exist a simple cycle through a given directed graph that visits all the nodes?'' To prove it is in NP, you need to show that a yes solution can be checked in polynomial time. In the above example, you can check to see if a given path through the graph is indeed a simple cycle in linear time. Therefore, the problem is in NP. You don't have to prove anything about the no solutions, and you don't have to prove anything about how you'd calculate a solution.

  2. Transform a known NP-Complete problem to this one in polynomial time. Suppose the problem in question is Q, and that L is a well-known NP-Complete problem like the 3-satisfiability problem. You need to show that if you have any instance of problem L, you can transform it into an instance of problem Q in polynomial time. Thus, if you could solve problem Q in polynomial time, you could solve problem L in polynomial time.
If you can do both of these things, then you have proved that a problem is NP-Complete. If you can prove that either of these things cannot be done, then you have proved that a problem is not NP-Complete. Sometimes you can't do come up with good proofs, and you just don't know.

The complexity classes P and NP-Hard may be put in terms of the above:

3-SAT - A Canonical NP-Complete Problem

3-SAT is a very simple NP-Complete problem. You are given a boolean expression, which is a big AND (∧) of clauses:

E = C0 ∧ C1 ∧ ... ∧ Cm-1

Each clause Ci is the OR (∨) of three literals, where a literal is either a variable xi or the negation of a variable ¬ xi. Here is an example with three clauses and four variables:

E = ( x0 ∨ ¬ x1 ∨ ¬ x2 ) ∧ ( x2 ∨ ¬ x1 ∨ ¬ x3 ) ∧ ( x3 ∨ ¬ x0 ∨ ¬ x2 ) ∧ ( ¬ x0 ∨ ¬ x1 ∨ ¬ x2 )

Given this definition, 3-SAT is simple -- is there an assignment of the variables so that E is true? In the above example, we can find such an assignment pretty easily: x0 and x3 are TRUE and x1 and x2 are FALSE. However, in general, 3-SAT can be a very difficult problem to solve. It's pretty easy to find a solution that is exponential in the number of variables. Call that number n. It is very easy to design a solution that is exponential in n: View each setting of the variables as an n-bit number, where bit i represents the assignment of variable x1. There are 2n of these numbers, and they represent all possible settings of the variables. So, enumerate them and test to see if E is true.

Now, it is an easy matter to prove that 3-SAT is in NP. How many different clauses can there be? Roughly (2n)3 (we'll go over that in class). That's a polynomial of n. If we have a solution, we can test its validity by simply setting the variables and seeing if E true. That test is polynomial time, so 3-SAT is in NP.

As for proving that 3-SAT is NP-Complete, that is well beyond the scope of this class. However, 3-SAT is a very popular problem for proving other problems NP-Complete.

How would we do that?

Suppose I have a problem, like Hamiltonian Path: Given a graph, can we find a path from one node to another that includes every other node in the graph exactly once? Here's how I'd use 3-SAT to prove it NP-Complete.

First, I'd prove it's in NP: If you give me a path, it is a trivial matter to test whether it is a Hamiltonian Path -- it's O(n) where n is the number of nodes in the path.

Next, what I'd do is figure out a way to convert 3-SAT to Hamiltonian Path. I would figure out a way to take a general expression E and turn it into a graph so that if I could solve Hamiltonian Path in polynomial time, then I could solve 3-SAT on E in polynomial time. I don't have to actually solve Hamiltonion Path -- I just have to show how I can use it to solve 3-SAT. By doing so, I have shown that Hamiltonian Path is NP-Complete.

Who Cares?

NP-Complete problems usually have easy-to-write exponential solutions. However, we cannot prove that they do not have polynomial time solutions. This is embodied in the equation:

P = NP?

It is a famous open question in theoretical computer science. Does its solution have practical worth? Maybe -- a lot of these problems pop up very naturally (doesn't Spellseeker from Lab B look a lot like Hamiltonian Path?), and if we could solve them in polynomial time rather than exponential, then that would be something!