What if P = NP?


Update (5 March): read a better version here.

I admit I now going slightly out of my depth, but I will try to explain what this is about and why it is interesting.

It is known that computers can solve some problems in what is called “polynomial time“: that is to say a finite time that is proportional to a polynomial of complexity of the input. The key thing is that these problems are computable using mechanical steps (ie algorithms) in a way that is (sometimes euphemistically) described as “quickly”.

These can be simple problems – like what is the sum of the first ten integers – or more complex ones, such as creating self-balancing trees, sorting database records alphabetically and so on.

These are the so-called “P” (for polynomial time class) problems. Here’s a definition:

A problem is assigned to the P (polynomial time) class if there exists at least one algorithm to solve that problem, such that the number of steps of the algorithm is bounded by a polynomial in n, where n is the length of the input.

Then there are another class of problems which seem fiendishly difficult to solve but which it is relatively simple to prove the correctness of any solution offered. These problems can also be solved (computed) in polynomial time – ie a finite time – and they can also be computed by a Turing machine (a simple model of a computer) and so an algorithmic solution exists. It is just that one cannot tell what that algorithm is. These are said to be solvable in unbounded polynomial time – and in the worst case the only way – it is thought – that a solution can be found is through an exhaustive search of all algorithms – in other words a form of “brute force“. These are the NP (Not in class P) problems.

Now most mathematicians think that NP does not equal P and that may or may not be a good thing as much of our internet commerce relies on encryption which is thought to be an NP problem.

(In Afghanistan in 2001 US cryptanalysts seemingly brute forced a Taliban Windows NT box but it used much weaker encryption than most electronic commerce.)

But what if it were the case that all seemingly NP problems were actually P problems? There are a lot of people studying this issue – but according to the New Scientist (their Christmas edition, the first in my subscription and delivered this morning, inspired me to write this) we should expect to wait at least until 2024 for an answer (by which point the problem – first formulated in 1971 – will have reached the age at which 50% of major mathematical problems will have been solved).

Some problems thought to be NP have already been shown to be P and there was a big fuss earlier in 2010 when a draft proof of P = NP (edit: it was actually P != NP) was published (the proof was flawed). And unlike, say, Fermat’s last theorem, proving P = NP is likely to have more or less immediate effects on the lives of many of us (how would you feel if you knew that it was possible, even if not likely, that someone had developed a way to quickly crack open all your internet credit card transactions?)

Of course, proving P = NP for all problems does not necessarily mean we will have determined polynomial time based solutions for all the current NP problems out there. But I would expect it would quickly lead to the solution of a multitude of them.

And, actually, I think the benefits to humanity would be enormous. The most likely immediate effects would be in improvements in computing/operating system efficiency. fast computers for less money and/or less energy consumption would be a huge benefit. From that alone many other benefits will flow in terms of information availability and universality.

4 thoughts on “What if P = NP?

  1. Pingback: No proof of P=NP after all (yet?) | cartesian product

  2. “These are the NP (Not in class P) problems.”

    NP stands for Non-Deterministic Polynomial. This class of problems has the characteristic that you can generate a potential solution, and test whether it is a solution in polynomial time. Therefore, if you had a non-deterministic turing machine (in the Theory of Computation sense), you could simultaneously try every possible solution, and find the answer in polynomial time.

  3. NP doesn’t mean “not in class P”, it’s a little more complicated than that. NP is the class of problems which can be solved by a Nondeterministic Turing machine in Polynomial time, or equivalently, the problems which can be verified by a deterministic Turing machine in polynomial time. P is in fact a subset of NP. Algorithms exist for many NP problems but all of them essentially require an exhaustive test of every possible solution, which takes non-polynomial time.

    Current encryption algorithms do not rely on P!=NP anyway, for the most part. If it is proven that P=NP, your credit card details aren’t at risk, but it may be the case that your computer runs faster.:)

  4. Pingback: P = NP once again | cartesian product

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