One of the most complex mathematical problems in the world is proving either that P ≠ NP or P=NP, a riddle that was first formulated in 1971 by mathematicians Leonid Levin and Stephen Cook. The question was one of seven millennium problems set by the Clay Mathematical Institute (CMI) in Cambridge, Massachusetts as being among the most difficult to solve.

Vinay Deolalikar, a computer scientist from Hewlett-Packard's research arm based in Palo Alto, California, claims to have solved the problem and stands to win $1 million if his proof is accepted.

P refers to problems with solutions that are easy to find, in other words P is a class of problems for which an answer can be found in polynomial time. NP refers to problems with solutions that are almost impossible to find, but easy to verify, or in other words a class of problems for which an answer can be verified in polynomial time.

The Clay Mathematical Institute offers the following problem to illustrate NP: you have a list of 400 university students to be accommodated but there is only room in the dormitory for 100 students. The Dean has provided a list of pairs of incompatible students who cannot be on the final list. This is an NP-problem because it is extremely difficult to go through all possible combinations to come up with a list, but very easy to verify once a list is obtained. According to CMI the number of possible combinations is greater than the number of atoms in the universe.

If P ≠ NP some complex problems, such as the one above, could never be solved efficiently and there are serious limits on what computers will be able to accomplish in the future, whereas if P = NP every such problem would have a polynomial time solution. Mr Deolalikar writes that this would have profound implications for applications such as cryptography and on the question of whether or not human creativity could be automated.

Among those skeptical of Deolalikar's claim is Scott Aaronson, assistant professor of electrical engineering and computer science at the Massachusetts Institute of Technology (MIT), who said on his blog that he will personally add $200,000 to the prize, even though he can ill afford it, and he has not read Deolalikar's entire paper. He said if the proof was correct it would change his life so dramatically that "having to pay $200,000 will be the least of it." He later also gave his reasons for being so confident in MIT's technology review.

Another skeptic is Professor Richard Lipton at Georgia Tech College of Computing, who has been working on the P vs NP problem for three decades.

Deolalikar's paper is available online, and is being peer-reviewed by computer scientists. To be considered for the million dollar prize the paper must be published in an approved mathematics publication and must also retain general acceptance by mathematicians two years after the proof is published.

**Explore further:**
P vs. NP -- The most notorious problem in theoretical computer science remains open

**More information:**
www.hpl.hp.com/personal/Vinay_ … r/Papers/pnp12pt.pdf

## plasticpower

## sender

## xponen

I think it is more plausible that P (is)!=NP, and our brain is just copying solution from the universe/its-environment... hence, our brain won't be original, the solution is a copy, but yet; the brain did solve some problem, minor one.

-For example; If I play a chess game and I'm good at it, then it is most probably that it is because I practice and I play by example (over my entire lifetime), rather than figuring out all the winning move all by myself, in my mind. I don't think even the most powerful computing substrate (like the brain) can ever figure out EVERYTHING by itself.

## gwrede

It is an interesting and entertaining read, even when one can't claim an adequate understanding of the text.

Personally, I'd be more at ease if P=NP is proved wrong than the opposite. I hope the paper is right.

## Doug_Huffman

## El_Nose

## gunslingor1

"The Clay Mathematical Institute offers the following problem to illustrate NP: you have a list of 400 university students to be accommodated but there is only room in the dormitory for 100 students. The Dean has provided a list of pairs of incompatible students who cannot be on the final list. This is an NP-problem because it is extremely difficult to go through all possible combinations to come up with a list, but very easy to verify once a list is obtained. According to CMI the number of possible combinations is greater than the number of atoms in the universe."

sounds like a mistatement of the problem. I don't think the final list would be infinite, how could it be, you only have room for 100 students?

## Valentiinro

## gunslingor1

Yeah, I realized that after posting. Problem is that I still think the example is unrepresentative of the problem. How many pairs could there really be that cannot be together? if it's anywhere close to 400!, then the list couldn't even have been generated.

Interesting though. This is my first exposer to this problem, though it does sound familiar.

I'm gonna study up before I look further like a moron.

## stealthc

That number could so be generated. At least a heck of alot easier than the recent refinement to pi that was calculated.

## xponen

## neoabraxas

## gunslingor1

there is a very simple algorithum I figured out (others probably figured this out as well) that will guarantee you never lose.

## Jovovic

http://blog.resea...mplexity

## dsa

That is what the guy is proving:

http://www.hpl.hp...lalikar/

http://www.hpl.hp...lalikar/Papers/pnp_synopsis.pdf

## abhishekbt

So, the solution will be something like 400! / 2! which comes out to be on the order of 3.2 x 10^868.

A quick Wikipedia check tell me that the estimated no. of atoms in the universe is in the order of 10^80.

Hope this helps realise the 'gravity' of the situation!

## Jarek

For example in 3SAT problem we have to valuate variables to fulfill all alternatives of triples of these variables or their negations – look that

(x OR y) can be changed into optimizing

((x-1)^2+y^2)((x-1)^2+(y-1)^2)(x^2+(y-1)^2)

and analogously seven terms for alternative of three variables. Finding global minimum of sum of such polynomials for all terms, would solve our problem. ( www.scienceforums...mputers/ )

It’s going out of standard combinatorial techniques to continuous world using gradient methods, local minims removing methods, evolutionary algorithms … it’s completely different realm: numerical analysis.

Such proof could work on concrete valuations, but here we work between them.

Does/can it cover also continuous approaches?

## ElasfarSovereign

## ElasfarSovereign

MIT's technology review.

## ElasfarSovereign

## Jarek

... and it looks new - there was needed huge amount of constrains earlier ... http://groups.goo...8?hl=en#

## gwrede

Try doing it standing in your hallway, with wife yelling, kids yelling, landlord sent thugs in your face, and a 2-week notice from the office.

So, do us a favor: solve the thing, get rich -- and as a side effect, help us out!

I'd call that "a favor to mankind, way surpassing those for which you get a Nobel Prize"!

## Jarek

More 'helpful' (and creating chaos...) would be a practical way to solve such problems - this numerical analysis translation could be a good start ... it has landed on my huge pile of problems I'm going to look closer sometime ...

But generally, there is also another way - just try to use physics, which have huge unexplored computability capability ... and it could give more options than quantum calculations, like for example:

http://www.electr...ter.html