3 ms·
Difficulty in cryptography is boolean. Is there an attack which has fewer steps than the brute force constant? If no? you’re good. If Yes? add more rounds, etc.
by zarzavat 5y ago
Difficulty in cryptography is boolean. Is there an attack which has fewer steps than the brute force constant? If no? you’re good. If Yes? add more rounds, etc.
- Ar-Curunir 5y agoNo, that’s not true. Plenty of cryptography is asymptotic.
- MauranKilom 5y agoBy that specific notion, any hash where you can shave a hair of complexity off the first round (e.g. via SAT) is insecure, because it takes ever-so-slightly less effort to brute force it that way than pure brute force. Adding more rounds doesn't change that.
- zarzavat 5y agoSure if you want to be pedantic. However even if you double the number of rounds, the number of operations required to brute force increases by only a factor of 2. Since the number of operations is likely to be a very large number already, such a change is insignificant and we can consider the brute force cost to be independent of the number of rounds which is usually a small integer.
- naniwaduni 5y agoThe only difference between 1 and 2^128 is 128 small constant factors of 2.