On Friday, Aug. 6, a mathematician at HP Labs named Vinay Deolalikar sent an e-mail to a host of other researchers with a 103-page attachment that purported to answer the most important outstanding question in computer science. That question is whether P = NP, and answering it will earn you $1 million from the Clay Mathematics Institute.

Last fall, MIT News published a fairly detailed explanation of what P = NP means. But roughly speaking, P is a set of relatively easy problems, NP is a set of incredibly hard problems, and if they’re equal, then a large number of computer science problems that seem to be incredibly hard are actually relatively easy. Problems in NP include the factoring of large numbers, the selection of an optimal route for a traveling salesman, and the so-called 3-SAT problem, in which you must find values that satisfy triplets of logical conditions. For instance, is it possible to contrive an attendance list for a party that satisfies the triplet (Alice OR Bob AND Carol) AND (David AND Ernie AND NOT Alice)? (Yes: Bob, Carol, David, and Ernie were there, but Alice wasn’t.) Interestingly, the related 2-SAT problem — which instead uses pairs of conditions, like (Alice OR Ernie) — is in the set of easy problems, P, as is the closely related XOR-SAT problem.

Most computer scientists believe that P doesn’t equal NP, and that’s what Deolalikar claimed to have proved. But by the following Monday, despite being on vacation in the Mediterranean and having time only to glance through the proof, MIT Associate Professor of Electrical Engineering and Computer Science Scott Aaronson had announced on his blog that he would mortgage his house and chip in another $200,000 if Deolalikar’s proof was correct. Last week, Aaronson took a few minutes to answer three questions about P and NP.

**Q. **Has the proof now been shown conclusively to be wrong?

**A. **I would say yeah. It was clear a couple days ago that there was a very serious gap in the statistical-physics part of the argument. It was not clear at all that the argument for showing why an NP-complete problem like 3-SAT was hard wouldn’t also show that problems like XOR-SAT are hard. Now, XOR-SAT is a variant of this satisfiability problem, which is known to be in P, which has an efficient solution. So if you’re proving that a problem is hard, but your proof could also be adapted to show that an easy problem is hard, then your proof must be fallacious: it proves too much. That’s the first check that people look for when someone announces a proof of P not equal to NP. Why doesn’t it also work for the easy problems?

The problem with saying that the thing has been conclusively refuted is that whenever anyone points to a problem like this, Vinay Deolalikar has been tending to respond, “Oh, yeah, sure, well, I’m going to address that in my next draft.” So it’s kind of a moving target. But I think it’s absolutely clear right now that at least the existing version does not solve the problem and furthermore wouldn’t solve the problem without some very, very major new ideas.

**Q. **Why were you so certain that there was a flaw in the proof?

**A. **P vs. NP is an absolutely enormous problem, and one way of seeing that is that there are already vastly, vastly easier questions that would be implied by P not equal to NP but that we already don’t know how to answer. So basically, if someone is claiming to prove P not equal to NP, then they’re sort of jumping 20 or 30 nontrivial steps beyond what we know today. So the first thing you look for is, What about steps one, two, and three? Can he explain even the easier questions, how he’s answering those? So I looked at the manuscript, and I didn’t see that.

The other check is the one that I already mentioned, which is, Why does the proof fail for variants of NP-complete problems which are known to be easy? What Deolalikar was doing is, he’s trying to argue that 3-SAT is hard by looking at its statistical properties. The problem is that 2-SAT and XOR-SAT, the problems that are easy, have very, very similar statistical properties, so it did not look like something that could distinguish the hard problems from the easy ones.

We have very strong reasons to believe that these problems cannot be solved without major — enormous — advances in human knowledge. So you look at the paper and you don’t see that it’s commensurate with the scale of the problem that it’s claiming to solve. This is not a problem that’s going to be solved by just combining or pushing around the ideas that we already have.

**Q. **Given that most people are pretty confident that P does not equal NP, what would the proof really do for us?

**A. **Yes, almost all of us believe already that P is not equal to NP. But this is one of those things where it’s not so much the destination as the journey. It’s the massive amount of new understanding of computation that’s going to be needed to prove such a statement. What are we trying to prove? That for solving all these natural optimization problems, or these search problems, or a proof of a theorem, finding the best schedule for airlines, breaking cryptographic codes — all these different things, that there’s no algorithm, no matter how clever, that’s going to solve them feasibly. So in order to prove such a thing, a prerequisite to it is to understand the space of all possible efficient algorithms. That is an unbelievably tall order. So the expectation is that on the way to proving such a thing, we’re going to learn an enormous amount about efficient algorithms, beyond what we already know, and very, very likely discover new algorithms that will likely have applications that we can’t even foresee right now.

Often in the history of theoretical computer science, the same ideas that you use to prove that something’s impossible can then be turned around to show that something else is possible, and vice versa. The simplest example of that is in cryptography, where you show some problem is hard to solve, and that gives you a code that is useful. But there are many other examples.

## Comments

Jon Allen

August 17, 2010

P=NP is not a meaningful determination without a qualification as to what exactly constitutes a problem, easy or hard. And the number of easy problems relative to the number of incredibly difficult problems does not necessarily have in itself significant meaning, without a compelling theory to back up this significance.

Rei

August 18, 2010

"Problems in NP include the factoring of large numbers, the selection of an optimal route for a traveling salesman..."

I suggest you read the wiki for more information

http://en.wikipedia.org/wiki/P...

Bhujang R

August 18, 2010

P=NP well could be possible in certain cases , for example the large factorial problem. This does conform to P=NP. However while in certain other problems it has no significant meaning. So the given theory is in-conclusive with the present order of algorithms.

allen mendelson

August 24, 2010

The universe can be described by maths.And all the maths can change to add. SO the universe can be described by a huge matrix. We assume this matrix is U. And we can use another matrix, C, to describe the movement of the universe. So the changed universe is C*U.

If we know the state of every particle in the universe, the randomness will not exist any more.

The universe is a running programme. Everything have been written. When we know what have been written, we are all deity.

George O.

October 26, 2010

It seems the basis for P=NP is on if the problem is scalable in a linear fashion. The more time you apply to it the greater the chances of solving it. This would work if you were trying to create more parking spaces in an n*m size parking lot that only allows for right turns. However for something that increases in complexity as it grows in size like tracking cellular mutation as a one cell organism multiplies would require a different approach.

In cases like that P=NP might not have to represent time applied to a problem but the number of processors (virtual or physical) that are used to solve the problem. The problem is broken down in to individual elements with each element handled by a different processor. In this case, the structure of the resulting network of processors may not need to be taken into consideration. It is merely an exercise to determine how much processing power is required to accomplish the task at hand.