2 ms·
Computational hardness doesn't depend on whether anybody knows an efficient algorithm. Just whether one exists.
by thomasahle 9y ago
Computational hardness doesn't depend on whether anybody knows an efficient algorithm. Just whether one exists.
- taeric 9y agoFair enough. I was obviously reading the above as having a polynomial time as having an efficient time one. Where efficient was shorthand for "quick." Both leaps, I concede were misguided.