In sci.math Randy <firstname.lastname@example.org> wrote: > I'm not a theorist, so my terminology will probably be a bit imprecise.
Unfortunately you are being very imprecise. Unfortunately this is an easy topic to be imprecise about, but one which you need to be precise about. You seem to be using a non-standard definition of NP-complete and NP-hard and as a result most of your answers are incorrect according to the standard definitions.
> But with luck, it'll help fill the gap between reality and theory...
> email@example.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 >> NP-Hard and NP-Complete?
> 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.
No, NP-Complete means that the problem is in NP and it is NP-hard. You can prove a problem is NP-hard by reducing it to another NP-hard problem. There is no need to use an NP-complete problem to show that a problem is NP-hard.
> 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.
No, we do not know that it is outside the class of P problems, nor do we know that the runtime is necessarily exponential.
> 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 NP-Hard >> 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 NP-Complete 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.
If a problem in NP-hard then we know that it is as hard as the hardest problems in NP. That is the definition of NP-hard. If you find a polynomial time solution for an NP-hard problem then P=NP. There is no maybe about it.
> 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 always-solvable-problems (all of NP) can be solved in > polynomial time.
No, there are plenty of problems "harder" than NP-complete problems that are solvable, assuming of course that P!=NP. For example, any NP-hard problem that is not NP-complete. Of course proving that a problem is not NP-complete is not trivial. I do not recall if there are any NP-hard problems that have been definitely proven to not be NP-complete.
> 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.
By definition an NP-hard problem is as hard as the hardest problem in NP. That is the definition of NP.
As I said earlier, your definitions of NP-complete and NP-hard are non-standard, and as a result most of your answers are incorrect according to the standard definitions.