3 ms·
You'd better double that bit count; there's a generalised algorithm which lets you break symmetric encryption of N bits in 2^(N/2) time. This probably still me
by Filligree 12y ago
You'd better double that bit count; there's a generalised algorithm which lets you break symmetric encryption of N bits in 2^(N/2) time.
This probably still means that 256 bits is good enough. Of course, then there's RSA to worry about..
- wahern 12y agoAre you thinking about the Birthday Paradox? Because a) that usually applies to hashes, and b) it applies when you're trying to find collisions by between two variable strings, not one fixed and one variable. Otherwise, you may want to get on the phone with everybody using AES-128, et al and let them know. 2^64 would be rather trivial to crack these days.
- wyager 12y agoHe's thinking of Grover's algorithm. And yet, if it could be implemented in a fast way, it would bring AES-128 into the range of trivial breakability.
- taejo 12y agoI think Filligree could have been referring to Grover's algorithm (requires many qubits of quantum computing power, which we don't know how to build, but we do know the algorithm)