P ≠NP: To begin with. There is (almost) no doubt whatever about that.
Why? To quote Scott Aaronson—an MIT complexity researcher interviewed in an excellent article on Technology Review about the background of the "P vs. NP" problem—if P = NP then, "we'd be living in a fundamentally different universe, and we'd probably have noticed by now."
Naturally, that leads to another, "Why?", which author John Pavlus answers cleanly and clearly:
"P versus NP" is more than just an abstract mathematical puzzle. It seeks to determine–once and for all–which kinds of problems can be solved by computers, and which kinds cannot. "P"-class problems are "easy" for computers to solve; that is, solutions to these problems can be computed in a reasonable amount of time compared to the complexity of the problem. Meanwhile, for "NP" problems, a solution might be very hard to find–perhaps requiring billions of years' worth of computation–but once found, it is easily checked.
The "P versus NP problem" asks whether these two classes are actually identical; that is, whether every NP problem is also a P problem. If P equals NP, every NP problem would contain a hidden shortcut, allowing computers to quickly find perfect solutions to them. But if P does not equal NP, then no such shortcuts exist, and computers' problem-solving powers will remain fundamentally and permanently limited. Practical experience overwhelmingly suggests that P does not equal NP. But until someone provides a sound mathematical proof, the validity of the assumption remains open to question.
That's why P vs. NP matters. And it's why P probably ≠NP. "But wait," you ask, "Wasn't this solved, like, a couple weeks ago."
Sadly, no. The proof offered by HP labs researcher Vinay Deolalikar isn't standing up very well against scrutiny and is not likely, at this point, to earn him the $1 million prize still up for grabs.
Technology Review: What does "P vs. NP" mean for the rest of us?
