2 ms·
No, ECDSA relies on the hardness of the discrete logarithm problem. Nothing to do with factoring, at least not in the classical sense. On a quantum computer,
by CamperBob2 1y ago
No, ECDSA relies on the hardness of the discrete logarithm problem. Nothing to do with factoring, at least not in the classical sense.
On a quantum computer, my understanding is that Shor's algorithm could potentially target both problems, though.
- cyberax 1y agoBoth systems are an example of a hidden Abelian subgroup problem. That is also why Shor's algorithm equally applies to both: https://en.m.wikipedia.org/wiki/Shor%27s_algorithm#Shor's_algorithm_for_discrete_logarithms https://en.m.wikipedia.org/wiki/Shor%27s_algorithm#Shor's_al... So a hypothetical classic algorithm that breaks the RSA is also highly likely to break the ECDSA.