3 ms·
Slightly tangential but how significant is O(sqrt(n)) speed up? Fast algorithms are slightly faster but intractable algorithms are still intractable?
by haecceity 5y ago
Slightly tangential but how significant is O(sqrt(n)) speed up? Fast algorithms are slightly faster but intractable algorithms are still intractable?
- eximius 5y agoIt makes memory, not compute, the limit. It halves the exponent on the number of operations. 2^128 - reasonably secure by todays standards! 2^64 - horribly insecure. (CAVEAT: where the speedup is applicable, which is often hard.)
- bawolff 5y ago2^64 is hardly horribly insecure. Its on the edge of what a gigantic compute cluster can do. So its not secure, but hardly horribly insecure. Especially even if you can get a quantum computter working, its not going to be on the same level of operations that a million dollars in AWS credits will get you. At least not for a very long time. Besides,in most places where that is an issue, its trivial to switch to 256bit algorithms > (CAVEAT: where the speedup is applicable, which is often hard.) Grover's algorithm has pretty wide applicability. Its the exponential speed ups like shor's algorithm that have super limited applicability.
- haecceity 5y agoI think the point is it lowers security by a factor of 2 in the exponent. Going from 2^64 to 2^63 lowers the amount of time to bruteforce a key by half.