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January 5, 2022

# A Plausible Reason It's So Hard To Prove P!=NP

Let’s do something that seems insane at first: let’s try to solve NP-complete decision problems using the “random” outputs of cryptographic hash functions.

It doesn’t matter exactly which hash functions we use. In fact, in this argument, the hash functions are really a stand-in for any polynomial-time algorithm, but I want to take advantage of your intuitive sense of what a cryptographic hash function is, so I’m calling them hash functions. The intuitive properties I want you to have in mind are that they are random-behaving and computationally irreducible, i.e. there’s no way to learn or predict properties of their output without actually computing them.

A bit of notation will help make things clear. Let H1, H2, … be any sequence of keyed polynomial-time hash functions where Hn takes a key k of size |k| = n as well as an arbitrary-length input x. For every key k, we can define the language HL(k) = { x | H|k|(k, x)’s first bit is 1 }. In other words, for every key of unbounded size, there’s a language HL(k), and to find out whether a string is in HL(k) you simply hash the string with the key k using the |k|-th algorithm and check if the first bit of the hash is 1.

Now, here’s the question: If we keep passing bigger and bigger keys to bigger and bigger members of the hash function family, is it possible that we would stumble upon some enormous key k such that HL(k) is an NP-complete language? If that happens, then P would equal NP. Although k would be impractically large (probably so large that its length would put even Graham’s number to shame), we could decide an NP-complete language in polynomial time using one call to Hn with the hard-coded key. So, could that happen?

There are infinitely many NP-complete languages, so in one sense there are infinitely many collision targets. Yet, on the other hand, it seems like we need an infinite sequence of ‘accidental’ bit collisions, one for every string, to collide an entire language.

If the hash functions were truly random, modeled as random oracles, the probability of a collision would work out to be exactly zero for each NP-complete language. But that’s a heuristic argument, there’s no real probabilities involved here! Although we are thinking about the hash functions as random-behaving, they’re not really random, they’re all deterministic computations. A language collision is therefore not ruled out by any statistical argument.

An information-theoretic argument can’t rule out a language collision either, because NP-complete languages have concise descriptions in the form of their polynomial-time solution-checking machine, and k can be arbitrarily large, holding arbitrary amounts of information.

If P is not equal to NP, then for any choice of hash function family, all infinitely-many, arbitrary-length keys k, HL(k) must definitely never collide with any NP-complete language. If P!=NP were true and unprovable, this wouldn’t be so weird, we’d just say that no language collision occurs, and that’s just a brute fact about math with no real need for a reason.

However, if P!=NP is true and provable, then this starts to look really weird. In this case it looks like whatever logic leads to P!=NP is somehow “forcing” hash functions’ “random” outputs to always miss the NP-complete languages, or vice-versa "forcing" the NP-complete languages to always miss the hash functions. The logic in the P!=NP proof would need to explain how these apparently-structureless (and arbitrary-size!) algorithms all have a "global" property of never colliding with an NP-complete language, or explain how the NP-complete languages 'conspire' to disagree with all of the hash functions.

If we had a proof of P!=NP, then we would know for sure that all hash functions’ outputs will always miss all of the NP-complete languages, without ever having to evaluate the hash functions! That seems really strange, maybe even as strange as a language collision occurring, depending on how much we believe in computational irreducibility.

If P!=NP is provable, then our intuitive notion of cryptographic functions behaving randomly, as well as the concept of computational irreducibility, become suspect. If we believe computational irreducibility is real, not just something we think is real because we are computationally bounded humans, we might be tempted to conclude that one of the following conjectures is true:

Conjecture 1: P=NP, but only by accident as described above, for a key so large that we’d never find out. (In this case, P=NP is probably unprovable, because even if we knew the key and it was small enough to write down, there's no reason to expect there to be a line of reasoning tying hash function outputs to the NP-complete language. It’s just an accident.)

Conjecture 2: P!=NP, but it’s unprovable, because the existence of a proof would violate our (admittedly informal) notions of computational irreducibility and cryptographic security.

But hold on, there are complexity classes known to be larger than P, like EXPTIME and classes containing undecidable languages. Why doesn’t this hash function collision trick apply there, too? Why don’t those proofs, the proof that P!=EXPTIME, and the proof that the halting problem is undecidable, both also show that all hash functions have the mysterious language-avoidance property that we’re worried violates computational irreducibility?

The answer can be found in the proofs of those theorems. Both the proof of P!=EXPTIME through the time hierarchy theorem and the proof that the halting problem is undecidable use diagonalization arguments. They define a language by way of an algorithm which would simulate the hash functions (in fact all ‘faster’ algorithms) on some inputs and disagree with them on purpose. Computational irreducibility is not violated by those results, because the proofs make reference to all of the hash functions, reference their evaluations, and then construct an algorithm to disagree with them. In those proofs, it’s not that the hash functions’ outputs miss the EXPTIME language or the undecidable language by accident, it’s exactly the other way around: the EXPTIME language or undecidable language make reference to, and were designed to miss, the hash functions.

Because of completeness, i.e. the fact that there are polynomial-time reductions between all NP-complete languages, it's not enough to construct one NP-complete language that misses the hash functions like in the proofs mentioned above. In fact, ALL NP-complete languages would have to miss ALL of the hash functions in order for P to not equal NP. It seems that it's either some cosmic accident that no collision occurs, or languages like SAT and TQBF (in the case of P vs PSPACE) are "secretly" diagonalizing, somehow, to disagree with all possible hash functions.

If this reasoning is correct, then it would explain why we can’t seem to improve our results any better than the time hierarchy theorem allows: irreducibility ensures that diagonalization is the only way to guarantee there isn't a collision. It would also explain why we can’t even rule out linear-time algorithms for NP-complete problems like SAT: good symmetric cryptography only needs linear time, so for all we know the hash function that produces the freak collision runs in linear time, too. Linear-time (or better) lower bounds for SAT or TQBF would count as evidence against this idea, since any proof of those results would explain to us exactly how at least some of the NP-complete or PSPACE-complete languages are conspiring to miss all of the linear-time hash functions, and linear-time hash functions should be just as computationally irreducible as hash functions with quadratic runtime or greater.

If instead of hash functions, we allowed arbitrary uniform families of polynomial-time functions, then HL(k) equalling an NP-complete language for some k would exactly be the definition of P=NP. By restricting ourselves to hash functions, where computational irreducibility makes some kind of intuitive sense, it’s easier to grasp how a proof of P!=NP either needs to do time-hierarchy-theorem-style diagonalization or exploit a lack of computational irreducibility present in all hash functions.

Of course this is just an intuitive argument. We haven’t proven anything here. Perhaps it’s possible to formalize some kind of “computational irreducibility hypothesis” and show that under that assumption, P!=NP is unprovable. That’s left as an exercise to the reader.

Personally, I think P versus NP is the wrong question to care about. If P=NP because of an impossible-to-discover collision then who cares? If P!=NP but we can find machine learning algorithms to solve small cases of "hard" problems then who cares? Studying concrete complexity, complexity theory without hiding constant factors in big-O notation and without hiding exponents in the word “polynomial”, seems like it would be more useful. If we could get good at that, then we could lower-bound the security of the cryptographic algorithms we actually use, and maybe even lower-bound the size of a freak-collision-producing key.

Concrete complexity is messy, though, because without big-O notation, the machine model matters, and we pretty much end up with a different “complexity theory” for every kind of machine. Unfortunately, I think computational irreducibility will—at least if my intuitive formulation of it is in the ballpark of what's true—prevent us from making much progress there, too.