Well, it's finally arrived the time to rethink the whole CA system.
A message from Comodo Hacker
41–44 of 44 posts
Re: A message from Comodo Hacker
#42Earlier quoted context omitted.
Hasn't everyone woken up some day and said "hey, I'm gonna break RSA today"? I guess that only happens when you have the experience of 1000 hackers...
LOL. Does this guy think he's really going to scare us with lines that could've come from a Michael Bay movie??
Aren't they just naively translated idioms which don't sound quite so hackneyed in Farsi? Dunno, just guessing.
Re: A message from Comodo Hacker
#43Earlier quoted context omitted.
No. Integer factorization is not NP-hard (so not NP-complete). (This isn't proven, but it's generally thought to be the case.) So, while doing a polynomial-time integer factorization would be hugely significant (and make all asymmetric encryption in the world useless), it would not prove P=NP.
> So, while doing a polynomial-time integer factorization would be hugely significant (and make all asymmetric encryption in the world useless) This is wrong in two ways. First, a polynomial-time algorithm could still be too slow to be practical, either because the degree of the polynomial were high or because the constant factor or asymptotically disappearing overhead were high. Second, discrete-logarithm-based cryp…
You're right, that statement was rather too broad. Thank you for the correction.
Re: A message from Comodo Hacker
#44Earlier quoted context omitted.
> I'll wait for the day when he announces he's broken it. And so will a massive part of the Computer Science/Mathematics/Physics community. Answering the P vs NP question is kind of a big deal. :D
No. Integer factorization is not NP-hard (so not NP-complete). (This isn't proven, but it's generally thought to be the case.) So, while doing a polynomial-time integer factorization would be hugely significant (and make all asymmetric encryption in the world useless), it would not prove P=NP.
I don't want to turn this into a complexity theory discussion thread but isn't it an NP problem? And does proving that it can be solved in polynomial time mean P == NP (and vice versa)?