126
There's a new way to break RSA that's faster than anything we've seen before (arstechnica.com)
submitted 1 month ago by to c/technology@lemmy.world
54 comments fedilink hide all child comments
you are viewing a single comment's thread
view the rest of the comments
[–] 7 points 1 month ago (4 children)

it relies on P vs NP

That is incorrect. It relies on perceived difficulty of factoring large semi-prime. P equals NP is related, of course, if it happens to be an NP-intermediate problem, but the difficulty of factoring is not because it's NP-complete.

There are already classical algorithms that are sub-exponential to solve the problem. If it were the case that the problem was NP-complete, then we would have had a big advance in our theory and practice based on it.

  • permalink
  • fedilink
  • source
  • parent
  • hideshow 4 child comments
    •
  • [–] -5 points 1 month ago (3 children)

    I'm not about to get into the weeds about P vs NP to explain a concept to someone that doesn't know much about encryption in the first place. I'm aware that factoring isn't really NP complete, but P vs NP is about solving vs verifying solutions for something in Polynomial time, which can't be done via brute force for RSA keys.

  • permalink
  • fedilink
  • source
  • parent
  • hideshow 3 child comments
    •
  • [–] 6 points 1 month ago (2 children)

    really NP complete

    I don't care who you want to get in the weeds with or what you wish to spar about, you're just not correct.

    There is no almost NP complete. There are strong and weak variants, sure, and those have particular implications.

    But you're drawing conclusions that don't exist from definitions that you clearly don't understand.

    Signed: Somebody who has taught theory of computation for over a decade.

  • permalink
  • fedilink
  • source
  • parent
  • hideshow 2 child comments
    •
  • [–] -2 points 1 month ago (1 child)

    If you want to be pedantic, fine. Integer factoring is not in P. Therefore the math used to encrypt RSA is not in P. The usage of integer factoring to encrypt therefore means it can't be decrypted in P using brute force. That's the point I'm making.

  • permalink
  • fedilink
  • source
  • parent
  • hideshow 1 child comment
    •
  • [–] 5 points 1 month ago

    We actually don't know if integer factorization is not in P, though.

    Right now, I think most of us would guess that it is a prime candidate for NP Intermediate. Hence why I mentioned it earlier.

    And you absolutely can solve it in polynomial time just not with classical architectures.

  • permalink
  • fedilink
  • source
  • parent