[eru]

> I don't even know what co-NP is. Could you explain?

See https://en.wikipedia.org/wiki/Co-NP That article even mentions integer factorisation.

> 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.

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

[keepamovin]

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).

[eru]

> Are there problems for which we can prove they are in NP but also prove they cannot have polynomial time (P time) solutions?

That's exactly the famous P!=NP question.

> 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??

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.

> 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.

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.)

[keepamovin]

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 ! :)