I proved a theorem and then "Linux guessed from it an efficient NP-complete algorithm" :-) and my computer broke, so I write in this chat instead of a normal communication. That was a joke, next there is a serious theorem:

I've proved that my algorithm is NP-complete under assumption that P=NP. bit.ly/2Q5Xv2V0 Formalize it!

I get a 404 for the bit.ly link :-(

https://math.portonvictor.org/2021/04/08/my-algorithm-that-is-np-complete-in-the-case-if-pnp/

It is simpler than you can imagine: proving provability of invertibility of an algorithm is clearly an NP-complete problem. To solve it I just enumerate and run parallel all algorithms.

Previous message had two typos, corrected.

I skimmed the paper, and I'm confused on the part about enumerating all algorithms—wouldn't this require that the set of all possible algorithms be countably infinite?

The set of all algorithms is obviously countably infinite.

Oh yeah, ignore me :-P

In that case, I guess my next question is: Can you interleave an infinite number of algorithms? (Wouldn't it take an infinite amount of time to finish e.g. the first step?)

https://www.google.com/search?q=markov+algorithms is simple to implement

Loop N algorithms adding one more algorithm after each loop.

Oh yeah, I remember that now (sorry, I don't remember much of the details from when I learned about complexity)

0 0 1 0 1 2 0 1 2 3...

So will somebody formalize my proof soon?

Just checked out the references in the paper and they pose a question that I'm also stuck on: For the algorithm to be in polynomial time, I assume each individual A_n also does. Is that requirement explicitly mentioned in the paper?

No, it's enough that just one "good" A is polynomial

Just one of An finishes normally, the rest are terminated in the middle.

BTW, my first solution attempt used (without me noticing that I have no definition) fractional number of algorithm steps. Then I was doing a sum of infinite number of steps being finite. That's not defined, but I think, it can be defined with some new concepts. But my new algorithm is simpler.

An is even not required to halt.

Okay I lied, I do have another question: How do we know when we've found the "correct" A_n? (More explicitly, does A_n return "yes"/"no", or does it return an actual proof of that fact? I think the latter, but I want to make sure I'm understanding it right.)

An "should" return a proof (or not a proof, in which case we ignore the result). Most results will be "garbage".

It returns a proof in some formal system like first-order predicate calculus. Maybe for coq implementation, it's easier to use a lambda calculus.

Makes sense. In that case, how do we know that a polynomial A exists? If A returns a proof of the correctness of an algorithm A', I don't *think* that A' being polynomial necessarily implies A is polynomial (cause proofs can get really big).

(Again, please pardon any errors I am making, will make, and/or have been making; I'm only a simple undergrad :-P)

So, the proof idea is that it searches for the first polynomial NP-complete algorithm and it produces the result. But it may terminate earlier if we find another algorithm that isn't polynomial or NP-complete but solves the problem for a particular input data.

But doesn't each algorithm we run during the interleaving step actually return a proof, rather than just "yes"/"no"?

We assume existence of some NP-complete (polynomial) algorithm by the conditions of my theorem.

I thought NP-complete only says "there exists a polynomial-time verifier" but not necessarily that the verifier returns a proof?

(in which case you have to know *a priori* which verifier is the right one)

On your first question. Proof size is polynomial because in our assumptions run time of the entire algorithm is polynomial

I use proofs as certificates. I specially specialized the problem for all certificates to be proofs.

An returns a proof. Verifier returns like yes/no

Is it valid to constrain the certificates like that? I was under the impression that NP-completeness only means 1) there exists some polynomial-length certificate as well as 2) some polynomial-time algorithm to check it (in which case the certificate itself doesn't need to be a proof)

Any finite object can be used as a certificate. Proofs are finite objects. So nothing invalid.

"A certificate can be a proof" doesn't necessarily imply that "All certificates are proofs" though, unless I'm getting lost somewhere

In general not all certificates are proofs. But in my article all certificates that I mention are proofs.

In other words, I replaced the problem of finding an NP-complete algorithm with equivalent problem of finding an NP-complete algorithm with proofs as certificates.

Couldn't it be true though that, for some NP-complete problem, no polynomial-length proof exists (instead, only a polynomial-length/-time certificate/certifier combination)?

I'm not quite sure. I think, you are right. But there is no need to consider all NP-complete problems (I consider just one).

Oh I see: Is it correct to say, then, that this decision algorithm only works for NP-complete problems which we know beforehand to have polynomial-length proofs?

No. My algorithm always solves my problem. The only question if it's polynomial. I proved it's polynomial if p=np.

I consider only one NP-complete problem

So is the overall statement, then, that there exists at least one NP-complete problem for which this algorithm is correct?

Yes, and moreover I proved it's polynomial if p=np

You are a Christian. I have a theological theory: the mind of Jesus is an NP complete algorithm

Victor Porton said:

Yes, and moreover I proved it's polynomial if p=np

Is P=NP even required for the overall algorithm to be polynomial, then? Couldn't the algorithm just search through all possible proofs even if P<>NP, since a polynomial-length proof must exist by the original assumptions?

If p!=np, then my algorithm isn't polynomial, as the inverse statement proves p=np

P=np if and only if my algorithm is polynomial

Suddenly I'm wondering: Doesn't the existence of this algorithm imply that the NP-complete problem we're solving isn't actually NP-complete?

because the algorithm, as far as I can tell, never actually uses the assumption that P=NP

meaning that we have a polynomial-time algorithm for solving this particular problem, which implies it's not actually NP-complete to begin with

I don't understand your logic. But I proved (unless some error) that my problem is NP-complete. The proof is that my problem subsumed proof finding.

Or, summarized: I think this algorithm might actually be a proof that an NP-complete problem cannot have a polynomial-length proof

Oh yeah, I've been implicitly assuming P<>NP

oops

So I guess it's more a general statement, then: Any problem with a polynomial-length proof is in P.

First, I don't know if my algorithm is polynomial. Second, p=np if and only if my algorithm is polynomial.

It's false

Correction: it's false unless p=np.

Will you work on formalizing this?

Looks like I almost proved p!=np. I will tweet.

I cannot tweet because my pc is broken. Here is my proposed p!=np proof:

Run my algorithms An in parallel not in equal time but each two times less CPU time than the previous.

Ignore this, my p!=np proof idea was erroneous.

I meant: your "might be actually a proof" is false.

FWIW: This sounds related to Levin search https://steemit.com/steemstem/@markgritter/leonid-levin-s-universal-algorithm, and especially it sounds very similar to the extension by Hutter https://arxiv.org/pdf/cs/0206022.pdf, which in addition proves optimality.

While Hutter does not discuss NP-completeness, since his algorithm is asymptotically optimal on each given problem, if P = NP it’d also be polynomial for NP problems. (EDIT: IIUC).

So, my discovery isn't as big as I thought?! It was known a constructive solution for p=np if p=np? Why do I see everywhere the claim that constructivity of p=np is unknown?!

Those are good questions, I am only pointing to potential related work in case that’s helpful.

Hutter does not discuss constructivity explicitly, and his proof seems a conventional proof in classical logic, not intuitionistic logic (I have no idea how hard it’d be to adapt it).

If I may also ask questions, I’m also confused about something, and I was unable to resolve my confusion by reading what your PDF:

Victor Porton said:

It is simpler than you can imagine: proving provability of invertibility of an algorithm is clearly an NP-complete problem. To solve it I just enumerate and run parallel all algorithms.

Why is that NP-complete, especially in an arbitrary formal system? It’s obvious that there are verifiable certificates for this problem, but that alone doesn’t restrict the problem to NP (let alone NP-complete); the certificate-checking must be polynomial in the size of the input.

In this case, the input is algorithm A, the certificate is a proof P that A is invertible, and the certificate-checker must check the proof against the formal system, in time polynomial in the size of A.

I don’t see why the size of P should be bound by the size of A, even informally.

No, the input of the algorithm, whose NP-completeness we check is X, A processes this input X. An is an intermediary data not the input data, maybe you confuse that A is an input of V. It's really so, but we don't check if V is NP-complete, so that's irrelevant.

Well, I wasn't using the symbols in the paper...

My point was just, it's not clear to me that "proving provability of invertibility of an algorithm is clearly an NP-complete problem"

in fact, I'm skeptical that problem is in NP: I don't think the size of the certificate is polynomial in the input algorithm (nevermind the proof of completeness)

See for instance the 2nd def. in https://en.wikipedia.org/wiki/NP_(complexity)#Formal_definition — the 2nd clause demands that the size of the certificate is polynomial.

To be more specific, the paper says this work applies to any formal system — but it's surprising if that applies to the standard presentation of first-order logic. That logic doesn't have cut, hence proof size is blown up exponentially (see e.g. https://en.wikipedia.org/wiki/Cut-elimination_theorem#Consequences_of_the_theorem and the linked https://scholar.google.com/scholar?hl=en&q=Don%27t+Eliminate+Cut+boolos&btnG=&lr=lang_de%7Clang_en%7Clang_it).

Paolo Giarrusso said:

in fact, I'm skeptical that problem is in NP: I don't think the size of the certificate is polynomial in the input algorithm (nevermind the proof of completeness)

It' spolynomial, because the NP-complete (by theorem assumption) algorithm B generates a polynomial-size certificate.

Is B is producing proofs in FOL without cut? Those can't be polynomial, so maybe some assumption is unsatisfiable (maybe like https://coq.zulipchat.com/#narrow/stream/237977-Coq-users/topic/Possible.20polynomial.20NP-complete.20solution/near/233896533, but I'm not sure if my question is on the same point)

- On the formalism, in fact, I already struggle seeing whether at page 1 that
`X is an NP ⇒ R(X)`

actually follows:`X in NP`

does not imply that`X`

produces a formal proof in our chosen formal system, while`R(X)`

does.

- also, I'm not convinced by the definition of NP. IIUC, you say that
`X is NP`

if and only if`∀ Y, X(Y) = Z ⇒ ∃polynomial-time algorithm X' : X' (Z) = Y.`

. However,`X in NP`

seems closer to`∃polynomial-time algorithm X': ∀ Y, X(Y) = Z ⇒ X' (Z) = Y.`

, which is strictly stronger.

On 1, the first problem seems that FOL certificates might be exponentially bigger than the actual certificates for NP completeness. The second is that R seems to demand X to produce not just a certificate, but a proof of correctness of itself — possibly on all inputs depending on the actual statement ("for every algorithm Y" means "∀ Y" but I can't guess where that goes, relative to all other quantifiers described informally)

Last updated: Sep 30 2023 at 06:01 UTC