Randy
Posts:
19
Registered:
12/13/04


Re: On NP Completeness and Intractability
Posted:
Oct 25, 2005 1:36 PM


I'm not a theorist, so my terminology will probably be a bit imprecise. But with luck, it'll help fill the gap between reality and theory...
nimmi_srivastav@yahoo.com wrote: > 1) All the literature pertaining to intractability that I have seen > refers to the Turing's Halting Problem. What are other common examples > of intractable problems?
Any task that is always solvable (decidable) and whose runtime grows exponentially with the size of its input set is NP Hard. In other words, if the input size of the task's data is N, the task's runtime is described by X^f(N), where X and f(N) are greater than 1 and are nondecreasing (they grow larger without ever growing smaller).
> 2) In somewhat layman terms, what exactly is the difference between > NPHard and NPComplete?
In lay terms, there is no difference. Both describe problems whose solutions require runtimes that grow exponentially as their input data size grows linearly. "NP Complete" simply indicates that it's an "NP Hard" problem that has been translated into another NP Hard problem that is already known to be NP Complete. In other words, the problem's runtime grows exponentially with size and 1) we know for sure that it lies within the class of NP problems and 2) OUTSIDE the class of P problems. The NPC problem lies within that subset of NP problems that we are certain they can never be solved in polynomial time, that is, unless it turns out that *ALL* problems in NP can be solved in polynomial time. Right now, it looks like P is smaller subset of problems that lie entirely within NP.
> 3) If a polynomial time algorithm is discovered for a hitherto NPHard > problem, does that have any implication for the P=NP question? > [I already know that if a polynomial time algorithm is discovered for a > hitherto NPComplete problem, then P=NP]
In general, yes. If a problem is NP Hard (exponential), then solving it in polynomial time is a good sign that: 1) P=NP, or 2) the problem isn't really exponential after all: either it's actually in P, or it's not as hard as the hardest problems in NP: the NPCs.
NP Complete just means that these are the hardest problems that we know that are always solvable. Solving any of these in polynomial time is a guarantee that all alwayssolvableproblems (all of NP) can be solved in polynomial time.
Thus, until we prove that a given NP Hard problem is equally as hard as the hardest problem in NP, we can't be sure that solving it in polynomial time guarantees that we can solve all other solvable problems in polynomial time.
> 4) When we refer to NP, without any qualifiers, does it imply NPHard?
No. NP includes all of P. So all problems that always can be solved in polynomial time (P) also are part of NP. Problems that can be solved, but not always in polynomial time are called "intractable", and almost certainly lie in NP. These problems are NP Hard. If a NP Hard problem ("A") can be shown to be translated polynomially into another problem that is known to be NPComplete, then we know that problem A DOES lie 1) in NP and 2) outside of P (it's never solvable in polynomial time), thus 3) we can confirm that it too is NP Complete.
Colloquially, any problem that is always solvable (decidable) is considered to be in NP. If it is always solvable in polynomial time, it's in P. If it's always solvable, but not in polynomial time, it's NP Hard. These problems appear to be harder than polynomial (outside of P), but we're not certain of that (they might be polynomial sometimes). NP Complete means we're certain that they're solvable but that the cost is always exponential.
Problems that are sometimes not solvable are outside P and NP. Such problems are called undecidable. They imply that some information that's essential to solve the problem is missing. However, undecidable problems are often solvable and even tractable if you can somehow find a way around the uncertainty. In effect, you solve a subset of the problem that's sufficient to serve your needs. Artificial Intelligence deals primarily in problems like these, as well as NP Hard problems. Operations Research attempts to find efficient methods to solve NP Hard problems, but also deals in uncertainty.
Formally, NP means that a "decision problem" (solution is "yes" or "no") can be solved in polynomial time using a nondeterministic turing machine (which do not exist). It also means that the problem's solution can be verified in polynomial time by a deterministic turing machine (which do exist).
For more on complexity theory:
http://en.wikipedia.org/wiki/NP_%28complexity%29
Randy
> > Thanks, > Nimmi >
 Randy Crawford http://www.ruf.rice.edu/~rand rand AT rice DOT edu

