Re:Quantum computers will break the encryption that protects...
by moefh on Saturday October 20, 2018 @07:16AM
(#18262729)
Elliptic curve crypto (the kind that is in common use, like ECDH) is just as broken by quantum computers as RSA.
The quantum algorithm that breaks RSA (Shor's algorithm) does it by efficiently solving the hidden subgroup problem[1] for finite Abelian groups. This can be used for factoring integers (which breaks RSA) and also for solving discrete logarithms (which breaks elliptic curve crypto).
Re:Quantum computers will break the encryption that protects...
by throwawaymath on Saturday October 20, 2018 @09:53AM
(#18263306)
The other answers have given good info, but speaking to your supposed factoring algorithm in particular - most of the internet's public key cryptography still relies on factoring problems. Your algorithm would be enormously useful and dangerous. You would likely be able to break anything that reduces to the hidden subgroup problem using it, which means you'd be breaking both elliptic curve and non-elliptic curve systems.
Re:Quantum computers will break the encryption that protects...
by ShaneCurran on Saturday October 20, 2018 @07:14AM
(#18262724)
Kind of. Elliptic Curve can be reduced to the Discrete Logarithm problem, which a variant of Shor’s Algorithm solves in polynomial time.
Re:Quantum computers will break the encryption that protects...
by wbhart on Saturday October 20, 2018 @08:19AM
(#18262916)
There are many applications of a fast factoring algorithm in number theory, and many people who would love to get their hands on one for that reason. You'd be exceptionally famous if you developed a very fast algorithm for this. That is to say nothing of the fact that it would say something about the fundamental complexity of various problems.
Re:Quantum computers will break the encryption that protects...
by the8472 on Saturday October 20, 2018 @06:41AM
(#18262667)
Certificates very much do use RSA. The ephemeral handshake commonly uses ECDH.
Elliptic Curve encryption for example does not depend (AFAIK) on the difficulty of factoring. (please tell me if this is wrong)