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Explaining computational complexity theory


Hoooo, doctoral comp flashback. Okay, here goes.

We start with the idea of a decision problem, a problem for which an algorithm can always answer “yes” or “no.” We also need the idea of two models of computer (Turing machine, really): deterministic and non-deterministic. A deterministic computer is the regular computer we always thinking of; a non-deterministic computer is one that is just like we’re used to except that is has unlimited parallelism, so that any time you come to a branch, you spawn a new “process” and examine both sides. Like Yogi Berra said, when you come to a fork in the road, you should take it.

A decision problem is in P if there is a known polynomial-time algorithm to get that answer. A decision problem is in NP if there is a known polynomial-time algorithm for a non-deterministic machine to get the answer.

Problems known to be in P are trivially in NP — the nondeterministic machine just never troubles itself to fork another process, and acts just like a deterministic one. There are problems that are known to be neither in P nor NP; a simple example is to enumerate all the bit vectors of length n. No matter what, that takes 2n steps.

(Strictly, a decision problem is in NP if a nodeterministic machine can arrive at an answer in poly-time, and a deterministic machine can verify that the solution is correct in poly time.)

But there are some problems which are known to be in NP for which no poly-time deterministic algorithm is known; in other words, we know they’re in NP, but don’t know if they’re in P. The traditional example is the decision-problem version of the Traveling Salesman Problem (decision-TSP): given the cities and distances, is there a route that covers all the cities, returning to the starting point, in less than x distance? It’s easy in a nondeterministic machine, because every time the nondeterministic traveling salesman comes to a fork in the road, he takes it: his clones head on to the next city they haven’t visited, and at the end they compare notes and see if any of the clones took less than x distance.

(Then, the exponentially many clones get to fight it out for which ones must be killed.)

It’s not known whether decision-TSP is in P: there’s no known poly-time solution, but there’s no proof such a solution doesn’t exist.

Now, one more concept: given decision problems P and Q, if an algorithm can transform a solution for P into a solution for Q in polynomial time, it’s said that Q is poly-time reducible (or just reducible) to P.

A problem is NP-complete if you can prove that (1) it’s in NP, and (2) show that it’s poly-time reducible to a problem already known to be NP-complete. (The hard part of that was provie the first example of an NP-complete problem: that was done by Steve Cook in Cook’s Theorem.)

So really, what it says is that if anyone ever finds a poly-time solution to one NP-complete problem, they’ve automatically got one for all the NP-complete problems; that will also mean that P=NP.

A problem is NP-hard if and only if it’s “at least as” hard as an NP-complete problem. The more conventional Traveling Salesman Problem of finding the shortest route is NP-hard, not strictly NP-complete.

Michael Sipser’s Introduction to the Theory of Computation is a great book, and is very readable. Another great resource is Scott Aaronson’s Great Ideas in Theoretical Computer Science course.

The formalism that is used is to look at decision problems (problems with a Yes/No answer, e.g. “does this graph have a Hamiltonian cycle”) as “languages” — sets of strings — inputs for which the answer is Yes. There is a formal notion of what a “computer” is (Turing machine), and a problem is in P if there is a polynomial time algorithm for deciding that problem (given an input string, say Yes or No) on a Turing machine.

A problem is in NP if it is checkable in polynomial time, i.e. if, for inputs where the answer is Yes, there is a (polynomial-size) certificate given which you can check that the answer is Yes in polynomial time. [E.g. given a Hamiltonian cycle as certificate, you can obviously check that it is one.]

It doesn’t say anything about how to find that certificate. Obviously, you can try “all possible certificates” but that can take exponential time; it is not clear whether you will always have to take more than polynomial time to decide Yes or No; this is the P vs NP question.

A problem is NP-hard if being able to solve that problem means being able to solve all problems in NP.

Also see this question:
What is an NP-complete in computer science?

But really, all these are probably only vague to you; it is worth taking the time to read e.g. Sipser’s book. It is a beautiful theory.

This is a comment on Charlie’s post.

A problem is NP-complete if you can prove that (1) it’s in NP, and
(2) show that it’s poly-time reducible to a problem already known to
be NP-complete.

There is a subtle error with the second condition. Actually, what you need to prove is that a known NP-complete problem (say Y) is polynomial-time reducible to this problem (let’s call it problem X).

The reasoning behind this manner of proof is that if you could reduce an NP-Complete problem to this problem and somehow manage to solve this problem in poly-time, then you’ve also succeeded in finding a poly-time solution to the NP-complete problem, which would be a remarkable (if not impossible) thing, since then you’ll have succeeded to resolve the long-standing P = NP problem.

Another way to look at this proof is consider it as using the the contra-positive proof technique, which essentially states that if Y –> X, then ~X –> ~Y. In other words, not being able to solve Y in polynomial time isn’t possible means not being to solve X in poly-time either. On the other hand, if you could solve X in poly-time, then you could solve Y in poly-time as well. Further, you could solve all problems that reduce to Y in poly-time as well by transitivity.

I hope my explanation above is clear enough. A good source is Chapter 8 of Algorithm Design by Kleinberg and Tardos or Chapter 34 of Cormen et al.

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