# A failed attempt to prove P == NP

In computer science, we have one really gigantic open question about complexity. In the lingo, we ask “Does P == NP?”. (I’ll explain what that means below.) On March 9th, a guy named Michael LaPlante posted a paper to ArXiv that purports to prove, once and for all, that P == NP. If this were the case, if Mr. LaPlante (I’m assuming Mr.; if someone knows differently, ie. that it should be Doctor, or Miss, please let me know!) had in fact proved that P==NP, it would be one of the most amazing events in computer science history. And it wouldn’t only be a theoretical triumph – it would have real, significant practical results! I can’t think of any mathematical proof that would be more exciting to me: I really, really wish that this would happen. But Mr. LaPlante’s proof is, sadly, wrong. Trivially wrong, in fact.

In order to understand what all of this means, why it matters, and where he went wrong, we need to take a step back, and briefly look at computational complexity, what P and NP mean, and what are the implications of P == NP? (Some parts of the discussion that follows are re-edited versions of sections of a very old post from 2007.)

Before we can get to the meat of this, which is talking about P versus NP, we need to talk about computational complexity. P and NP are complexity classes of problems – that is, groups of problems that have similar bounds on their performance.

When we look at a computation, one of the things we want to know is: “How long will this take?”. A specific concrete answer to that depends on all sorts of factors – the speed of your computer, the particular programming language you use to run the program, etc. But independent of those, there’s a basic factor that describes something important about how long a computation will take – the algorithm itself fundamental requires some minimum number of operations. Computational complexity is an abstract method of describing how many operations a computation will take, expressed in terms of the size or magnitude of the input.

For example: let’s take a look at insertion sort. Here’s some pseudocode for insertion sort.

def insertion_sort(lst):
result = []
for i in lst:
for j in result:
if i < j:
insert i into result before j
if i wasn't inserted, add it to the end of result
return result


This is, perhaps, the simplest sorting algorithm to understand - most of us figured it out on our own in school, when we had an assignment to alphebetize a list of words. You take the elements of the list to be sorted one at a time; then you figure out where in the list they belong, and insert them.

In the worst possible case, how long does this take?

1. Inserting the first element requires 0 comparisons: just stick it into the list.
2. Inserting the second element takes exactly one comparison: it needs to be compared to the one element in the result list, to determine whether it goes before or after it.
3. Inserting the third element could take either one or two comparisons. (If it's smaller than the first element of the result list, then it can be inserted in front without any more comparisons; otherwise, it needs to be compared against the second element of the result list. So in the worst case, it takes 2 comparisons.
4. In general, for the Nth element of the list, it will take at most n-1 comparisons.

So, in the worst case, it's going to take 0 + 1 + 2 + ... + n-1 comparisons to produce a sorted list of N elements. There's a nice shorthand for computing that series: $\frac{(n-1)(n-2)}{2}$, which simplifies to \frac{n^2 -3n + 2}{2}, which is O(n2).

So while we can't say "computing a list of 100 elements will take 2.3 seconds" (because that depends on a ton of factors - the specific implementation of the code, the programming language, the machine it's running on, etc.), we can say that the time it takes to run increase roughly proportionally to the square of the size of the input - which is what it means when we say that insertion sort is O(n2).

That's the complexity of the insert sort algorithm. When we talk about complexity, we can talk about two different kinds of complexity: the complexity of an algorithm, and the complexity of a problem. The complexity of an algorithm is a measure of how many steps the algorithm takes to execute on an input of a particular size. It's specific to the algorithm, that is, the specific method used to solve the the problem. The complexity of the problem is a bound that bounds the best case of the complexity of any possible algorithm that can solve that problem.

For example, when you look at sort, you can say that there's a minimum number of steps that's needed to compute the correct sorted order of the list. In fact, you can prove that to sort a list of elements, you absolutely require $n lg n$ bits of information: there's no possible way to be sure you have the list in sorted order with less information that that. If you're using an algorithm that puts things into sorted order by comparing values, that means that you absolutely must do O(n lg n) comparisons, because each comparison gives you one bit of information. That means that sorting is an O(n log n) problem. We don't need to know which algorithm you're thinking about - it doesn't matter. There is no possible comparison-based sorting algorithm that takes less than $O(n \log n)$ steps. (It's worth noting that there's some weasel-words in there: there are some theoretical algorithms that can sort in less than O(n lg n), but they do it by using algorithms that aren't based on binary comparisons that yield one bit of information.)

We like to describe problems by their complexity in that way when we can. But it's very difficult. We're very good at finding upper bounds: that is, we can in general come up with ways of saying "the execution time will be less than O(something)", but we are very bad at finding ways to prove that "the minimum amount of time needed to solve this problem is O(something)". That distinction, between the upper bound (maximum time needed to solve a problem), and lower bound (minimum time needed to solve a problem) is the basic root of the P == NP question.

When we're talking about the complexity of problems, we can categorize them into complexity classes. There are problems that are O(1), which means that they're constant time, independent of the size of the input. There are linear time problems, which can be solved in time proportional to the size of the input. More broadly, there are two basic categories that we care about: P and NP.

P is the collection of problems that can be solved in polynomial time. That means that in the big-O notation for the complexity, the expression inside the parens is a polynomial: the exponents are all fixed values. Speaking very roughly, the problems in P are the problems that we can at least hope to solve with a program running on a real computer.

NP is the collection of problems that can be solved in non-deterministic polynomial time. We'll just gloss over the "non-deterministic" part, and say that for a problem in NP, we don't know of a polynomial time algorithm for producing a solution, but given a solution, we can check if it's correct in polynomial time. For problems in NP, the best solutions we know of have worst-case bounds that are exponential - that is, the expression inside of the parens of the O(...) has an exponent containing the size of the problem.

NP problems are things that we can't solve perfectly with a real computer. The real solutions take an amount of time that's exponential in the size of their inputs. Tripling the size of the problem increases its execution time by a factor of 27; quadrupling the input size increases execution time by at least a factor of 256; increasing the input by a factor of 10 increases execution time by at least a factor of 10,000,000,000. For NP problems, we're currently stuck using heuristics - shortcuts that will quickly produce a good guess at the real solution, but which will sometimes be wrong.

NP problems are, sadly, very common in the real world. For one example, there's a classic problem called the travelling salesman. Suppose you've got a door-to-door vacuum cleaner salesman. His territory has 15 cities. You want to find the best route from his house to those 15 cities, and back to his house. Finding that solution isn't just important from a theoretical point of view: the time that the salesman spends driving has a real-world cost! We don't know how to quickly produce the ideal path.

The big problem with NP is that we don't know lower bounds for anything in it. That means that while we know of slow algorithms for finding the solution to problems in NP, we don't know if those algorithms are actually the best. It's possible that there's a fast solution - a solution in polynomial time which will give the correct answer. Many people who study computational complexity believe that if you can check a solution in polynomial time, then computing a solution should also be polynomial time with a higher-order polynomial. (That is, they believe that there should be some sort of bound like "the time to find a solution is no more than the cube of the time to check a solution".) But so far, no one has been able to actually prove a relationship like that.

When you look at NP problems, some of them have a special, amazing property called NP completeness. If you could come up with a polynomial time solution for any single NP-complete problem, then you'd also discover exactly how to come up with a polynomial time solution for every other problem in NP..

In Mr. LaPlante's paper, he claims to have implemented a polynomial time solution to a problem called the maximum clique problem. Maximum clique is NP complete - so if you could find a P-time solution to it, you'd have proven that P == NP, and that there are polynomial time solutions to all NP problems.

The problem that Mr. LaPlante looked at is the maximal clique problem:

• Given:
1. a set of $V$ atomic objects called vertices;
2. a set of $E$ of objects called edges, where each edge is an unordered pair $(x, y)$, where $x$ and $y$ are vertices.
• Find:
• The largest set of vertices C=$\{v_1, ..., v_n\}$ where for any $v_i$, there is an edge between $v_i$ to every other vertex in $C$.

Less formally: given a bunch of dots, where some of the dots are connected by lines, find the largest set of dots where every dot in the set is connected to every other dot in the set.

The author claims to have come up with a simple P-time solution to that.

The catch? He's wrong. His solution isn't P-time. It's sloppy work.

His algorithm is pretty easy to understand. Each vertex has a finite set of edges connecting it to its neighbors. You have each node in the graph send its list of its neighbors to its neighbors. With that information, each node knows what 3-cliques its a part of. Every clique of size larger than 3 is made up of overlapping 3-cliques - so you can have the cliques merge themselves into ever larger cliques.

If you look at this, it's still basically considering every possible clique. But His "analysis" of the complexity of his algorithm is so shallow and vague that it's easy to get things wrong. It's a pretty typical example of a sloppy analysis. Complexity analysis is hard, and it's very easy to get wrong. I don't want to be too hard on Mr. LaPlante, because it's an extremely easy mistake to make. Analyzing algorithmic complexity needs to be done in a careful, exacting, meticulous way - and while Mr. LaPlante didn't do that, most people who are professional programmers could easily make a similar mistake! But the ultimate sloppiness of it is that he never bothers to finish computing the complexity. He makes vague hand-wavy motions at showing the complexity of certain phases of his algorithm, but he never even bothers to combine them and come up with an estimate of the full upper-bound of his algorithm!

I'm not going to go into great detail about this. Instead, I'll refer you to a really excellent paper by Patrick Prosser, which looks at a series of algorithms that compute exact solutions to the maximum clique problem, and how they're analyzed. Compare their analysis to Mr. LaPlante's, and you'll see quite clearly how sloppy LaPlante was. I'll give you a hint about one thing LaPlante got wrong: he's taking some steps that take significant work, and treating them as if they were constant time.

But we don't even really need to look at the analysis. Mr. LaPlante provides an implementation of his supposedly P-time algorithm. He should be able to show us execution times for various randomly generated graphs, and show how that time grows as the size of the graph grows, right? I mean, if you're making claims about something like this, and you've got real code, you'll show your experimental verification as well as your theoretical analysis, right?

Nope. He doesn't. And I consider that to be a really, really serious problem. He's claiming to have reduced an NP-complete problem to a small-polynomial complexity: where are the numbers?

I'll give you a good guess about the answer: the algorithm doesn't complete in a reasonable amount of time for moderately large graphs. You could argue that even if it's polynomial time, you're looking at exponents that are no smaller than 3 (exactly what he claims the bound to be is hard to determine, since he never bothers to finish the analysis!) - a cubic algorithm on a large graph takes a very long time. But... not bothering to show any runtime data? Nothing at all? That's ridiculous. If you look at the Prosser paper above, he manages to give actual concrete measurements of the exponential time algorithms. LaPlante didn't bother to do that. And I can only conclude that he couldn't gather actual numbers to support his idea.

# Holy Freaking Cow… P != NP??

Very big, very exciting news in the theoretical comp sci world today!

A group at HP research has published a proof that, if correct, shows that the classic problem of computational complexity has been solved, once and for all. It’s still far from certain that it’s correct. But it’s the most credible attempt that I’ve ever seen, and it’s getting some preliminary favorable feedback from big-names in complexity theory. In fact, one of the biggest names in complexity theory, Stephen Cook, has apparently said that it should be taken seriously. If Cook thinks it’s credible, then it’s damn-well credible. It might not be correct, but it’s not just crackpottery either.

For the non-CS folks out there, here’s my attempt to explain what the problem is, and what the solution means.