2 ms·
Yes, it would. The adversary in cryptography is assumed to be worst-case evil; it'll do whatever it can to break your algorithm, as long as it's performing a po
by Ar-Curunir 9y ago
Yes, it would. The adversary in cryptography is assumed to be worst-case evil; it'll do whatever it can to break your algorithm, as long as it's performing a polynomial amount of computation. Remember, you're not just fighting your friendly neighborhood hacker, but possibly entire nation-states.
- PeterisP 9y agoYou can easily have crypto that can't be broken with all the resources and time in the observable universe even if it's "just" polynomial with a sufficiently high coefficient.
- kevincox 9y agoYes, but it becomes significantly more expensive. It might work but it is a huge difference in cost tradeoffs.