A simple explanation to P and NP

[andrewducker]

lebeautemps asked for a simple explanation of the whole P=NP thing that's been circulating the internet recently. I figured I'd give it a go - I'd appreciate it if the more knowledgable on my friends list could correct anything particularly stupid I'd said...

A "P" problem is one that is easily solved through a series of steps, without trying every single combination - such as "sort a bunch of names into alphabetical order". You don't have to try every single possible list of names in order to sort a list of names, you can just use a series of comparisons to shuffle them up and down until they're sorted.

An non-P problem is one that has no known efficient set of steps for producing an answer, and thus requires you to try possibilities until you find one that's right. A common example of one that is thought to be hard is The Travelling Salesman problem - "given a bunch of cities and a bunch of roads connecting them, what is the shortest route that will take the salesman to each city exactly once?" There's no known solution to this problem other than "start trying solutions and keep going until you've found one." (although there are ways of excluding obviously wrong answers quickly).

NP is the superset of all problems that are easy to check, P is the smaller subset of all problems that are easy to solve, and proving that they are the same thing would mean that all problems that are easy to check are also easy to solve.

One of the reasons this is important is that pretty much all of the security methods we use online rely on things like "integer factorization" not being P - the fact that we can multiply two large numbers together to get an answer very quickly, but breaking that large number back into the two component parts requires every possibility to be checked (which takes years/decades for very large numbers).

The recent fuss is because of a paper that was published claiming that P!=NP - i.e. that there are definitely problems that are easy to check, but not easy to solve. This would mean, for example, that there is no simple way to convert the big number back into its two components, and thus that all of our security is safe (from that particular direction, anyway).

Comments

[call-waiting.livejournal.com]

I should point out is that the proof (if it's valid) is that P != NP, which is far less interesting than P = NP...

[ciphergoth.livejournal.com]

An "NP" problem is one that has no known set of steps for producing an answer

No, an NP problem is any problem where you can efficiently check the answer. All problems in P are in NP, for example. What's known about a problem doesn't affect what complexity class it's in.

[simont]

An "NP" problem is one that has no known set of steps for producing an answer, and thus requires you to check every possibility until you find one that's right.

Doing that is a set of steps for producing an answer! You probably meant to say "no efficient set of steps" here, or some such.

Also, technically, NP is a superset of P – there's no implication that "an NP problem" doesn't have a polynomial-time solution. Colloquial usage is to only describe problems as NP if they aren't (known to be) in P, on the pragmatic basis that if they were easier than that you'd have said so, but that's not what it really means.

I think the major thing I would add to this explanation is a paragraph explaining why the question is open, along the lines of:

At first sight it seems obvious that NP is a bigger class of problems than P, or in other words, that when a problem looks as if it needs you to go through all the possibilities one by one it's likely that there is in fact no quicker way to solve it. However, it turns out that this is surprisingly hard to prove (or disprove), so there's still a possibility in principle that the classes P and NP might turn out to be identical – which would mean that any problem for which you could efficiently check a candidate solution would also have an efficient way of finding a solution from scratch. Most computer scientists believe this is not true, and that P and NP are just as different as they appear, but a proof would be big news either way. (Though it would be bigger news if it went the surprising way and said that P and NP were equal.)

[ciphergoth.livejournal.com]

Also lots and lots of public-key cryptography doesn't depend on integer factorization being hard - other important hard problems include the discrete logarithm problem in Z*_p, or in a group based on elliptic curves, or the computational Diffie-Hellman problem or the Decisional Diffie-Hellman problem or... lots of other things.

And in cryptography, it's not worst-case hardness that matters - it's average case hardness.

If P turns out to equal NP, cryptography will have big problems - we'll be on the search for problems where the polynomial describing their hardness has a big exponent - but if we prove P != NP we'll still be a long way from having a cryptosystem we can prove good things about without making any assumptions about hardness.

[andrewhickey.livejournal.com]

That's about as good an explanation as I've seen without getting into the technical details. You might, however, want to use a paraphrase of this explanation as to why it's important ( from http://www.scottaaronson.com/democritus/lec6.html - the best *technical* explanation I've seen, from a wonderful series of lectures on quantum computing):

"Dude. One way to measure P vs. NP's importance is this. If NP problems were feasible, then mathematical creativity could be automated. The ability to check a proof would entail the ability to find one. Every Apple II, every Commodore, would have the reasoning power of Archimedes or Gauss. So by just programming your computer and letting it run, presumably you could immediately solve not only P vs. NP, but also the other six Clay problems. (Or five, now that Poincaré is down.) "