21 replies
6d15h
There'll probably be discovered a class of problems which is hard and Q hard, provably and we'll be set. Something basically new to cryptography, that was tried once in the past but failed until seen through a new light of more recent maths or research fashion.
But until then it seems to me like something based on the difficulty of attacking hash functions would be a good bet for Q resistant. Totally unsure how to make a PK scheme from that, but it has a few nice properties:
- hashes are often tuneable, you can add more state and increase the keyspace/security
- good hashes don't have any weaknesses that Q can exploit
- hashes are often pretty fast
- hashes are well studied
- hashes seem to be hard in C and hard in Q
We don't even know the complexity class of factorization or discrete log, yet we still use those problems in DH, RSA, ECDSA, ...
All of those problems are known to be in NP and co-NP. In that sense, we know some complexity classes they belong to.
However, we don't know if these bounds are tight, or whether they are eg in P, or something in between.
We don't know that factorization is NP-complete> Show me a reduction from SAT to factorization.
It's kind of trivial to say it's in NP because we can verify in P time, that's not a criticism of you just of the definition!!
I think a better definition of NP is "only nonpoly algos can exist, no P algos can exist". By that definition of NP, we don't even know that it's in NP strictly because there could exist P algorithms for solving it. It's more in 'unknown-NP' if that were a class! hahaha! :)
Yes? No one ever said it was.
None of the common cryptographic problems are expected to be NP-complete, even if they aren't in P. That's because they are known to be in both NP and in co-NP, and it's expected that NP != co-NP.
In what sense is that a 'better' definition than the standard definition? It sounds like what you are talking about is NP\P (where \ is set subtraction, ie 'NP minus P').
I think some people have asked whether it was. I'm not saying you did, just thought it was interesting! Haha :)
I don't even know what co-NP is. Could you explain?
I think that's a better definition because I find it more predictive and useful to think about: pretty concrete to know that you can't have a polytime algo for it.
Yeah, I guess what you're saying about NP\P is right in that it's a restatement of the definition of what I said, haha! I'm not an expert this is just what I think :)
See https://en.wikipedia.org/wiki/Co-NP That article even mentions integer factorisation.
Well, that's a non-standard definition for NP, and you would have a hard time talking to anyone. And at the moment we have no clue whether your 'NP' has any problems in it at at all, or whether it's an empty set. In that sense, it's a very impractical definition.
Btw, there's some nice alternative but equivalent definitions for traditional NP. The classic definition is basically, NP are those problem that you can check in polynomial time if someone gives you a hint (ie they give you the answer and whatever else you need, but you need to verify, you can't trust the hint.)
A nice alternative definition says that with access to randomness, that hint needs to be at most O(log n) long, and you also only need to even look at 3 randomly chosen bits of that short hint, and you are still guaranteed to suss out any fake answer with at least 66% probability. See https://en.wikipedia.org/wiki/PCP_theorem
Thanks for the alt NP definition. I'd be fine to talk to people we just have to clarify the definitions first. Haha! :) I think mine's good but I get if you differ, no worries.
It's actually a very fascinating definition and question: Are there problems for which we can prove they are in NP but also prove they cannot have polynomial time (P time) solutions?
I did check out that wiki page first, but found it super difficult to parse. Do you have some insight that could help me understand more simply/intuitively??
For instance, I found the definition of NP as P if you have an NFA, to be super easy to understand. But when that wiki starts talking about "certificates" I just have no idea.
That is, co-NP is the set of decision problems where there exists a polynomial {\displaystyle p(n)} and a polynomial-time bounded Turing machine M such that for every instance x, x is a no-instance if and only if: for some possible certificate c of length bounded by {\displaystyle p(n)}, the Turing machine M accepts the pair (x, c).
That's exactly the famous P!=NP question.
Scott Aaronson might have some good intro material on his blog. Otherwise, you can just ask your favourite search engine (or AI bot) for some intro material.
The certificate is the 'cheatsheet' or 'hint'. Basically the question is, how well can you do in an exam where you have to show your work, if someone gives you all the answers? (But that guy is a troll, so you can't trust him, and still need to verify everything.)
Cool, thank you. Yeah that makes sense. I didn't expect you to actually explain the entire thing, I just wondered if you had some, you know, insight. It's all good hahaha! :) I like your cheatsheet, I guess that applies to your previos definition of co-NP ! :)
I think this what alot of people get wrong. "N' in NP does not stand for "not" it stands for "non-deterministic". Meaning you can solve in P time with a non-deterministic Turing machine, or alternatively, a function executing on all inputs in parallel.
So maybe it should really be P and NDP.
That's a good explanation. I didn't know that.
I like to explain non-determinism in terms of getting a hint, or having an (untrusted) cheatsheet in a test. Or always making lucky guesses (but you don't trust your guesses).
But as long as your parallel executions don't interact at all, the definitions are identical, I think.
I always found that part odd. I’d assume you would want the problem you build your crypto system built around to be NP-complete, since that would seem to put you on the firmest possible ground. And yet those are most likely not NP-complete, and I think the post-quantum systems proposed aren’t NP complete either.
Maybe being NP-complete isn’t as important as I realize? Or maybe there’s something about NP-complete problems that make them less amenable to be a valid crypto system?
No crypto-problem is NP-complete. People tried that for a while, see https://en.wikipedia.org/wiki/Knapsack_cryptosystems but it didn't work.
To simplify a bit, the problem is that to work as a crypto system your particular problems needs to be both in NP and in co-NP. And we know of no problem that is both NP-complete and in co-NP. It's widely conjectured that there is no such problem. See https://en.wikipedia.org/wiki/Co-NP that page even mentions integer factorisation.
That's why you can't just take the NP-complete problem itself as a basis for your cryptosystem, you have to pick some subset of instances that's also in co-NP. And apparently it's almost impossible for us to pick such a subset, but still have the instances be hard enough to solve on average.
Just remember that one time pads are not only one of the simplest encryption schemes they are proven secure in any quantum computing regime. It's a shame about the key transport problem.
One of my favorite sci-fi macguffians, was in the book "Fire upon the deep" where they were space truckers with a cargo of one time pad keys.
This is also true of most symmetric cryptography algorithms. (Of which one-time pad is one of. So is AES-256. It's more convenient to just use AES-256.)
The implied context of "post-quantum cryptography" is usually asymmetric cryptography.
Check out the Winternitz One-Time Signature
Sphere10.com/articles/cryptography/pqc/wots
Signing many things with one identity is possible by precomputing a Merkle tree, but this takes time and the signatures get big.
SPHINCS+ is a complete signature scheme that carries that idea to it's completion. Shameless plug for where I explain it: https://er4hn.info/blog/2023.12.16-sphincs_plus-step-by-step...
https://en.wikipedia.org/wiki/McEliece_cryptosystem
Yes, it's reasonably fast even at very long lengths, but the main problem is the very long lengths. Code based and not hash based though.
Edit: not very fast to generate a key though. It's mostly used for non-ephemeral stuff.
The new algorithm purports to solve LWE with certain choices of parameters. LWE is the problem of solving a linear system of equations over a finite ring, where each equation has an additive error from a certain distribution.
McEliece has a public key that is a general linear code. A code is a bunch of linear equations constraining codewords, and codewords are vectors over a finite field, and decoding a code is solving those equations subject to errors from a given distribution. Sounds familiar?
They’re not the same problem, and the distributions are different in rather fundamental ways (which may or may not make a difference), but they are quite related. I would not move my eggs to the McEliece basket right now.
Hash-based signatures sound as safe as ever.
There are a bunch of hash based signature schemes, e.g., SPHINCS https://sphincs.org/