New answers tagged

1

Shamir secret sharing scheme is one of the examples.


0

Bitcoin blockchain cryptography algorithm is based on hash function. Finding next coin/block in block chain is equivalent to finding next payload resulting in hash having sufficient amount of leading zeros (adjusting difficult y is based on adjusting value of amount of zeros)


0

I do not agree to all of the combination advantages. All common memory-hard function are at the same time CPU-hard. And I'm not aware of an entropy reducing Password-based KDF in normal scenarios (of course any maximum password length theoretically reduces the entropy for larger pass phrases). There are also disadvantages: From the point of view of ...


2

This is homework, and so I won't give you the answer outright; I will give hints: Hint 1: how could you efficiently generate a random pair $x_1, y_1$ with $x_1^e = y_1$, without resorting to the Oracle? Hint 2: how could you use the above observation to accelerate your algorithm?


12

You have clarified the question as asking about whether replacing ShiftRows with a random byte permutation would strengthen AES against differential attacks. It would not. ShiftRows and MixColumns were carefully selected to work in tandem, such that every byte affects every other byte in the state within just two rounds. MixColumns ensures that every ...


7

I assume that you mean the S-box. The answer is NO! Randomly chosen S-boxes are not good choices for differential and linear cryptanalysis. When Biham and Shamir presented differential attacks on DES, one of the things that they showed was that if you replace the S-boxes in DES with randomly chosen ones, then the differential attack becomes much more ...


0

Sure. $\langle\{s_1,s_2,s_3,...,s_n\},\{\text{selected}_{\hspace{.03 in}1},\text{selected}_{\hspace{.04 in}2},\text{selected}_{\hspace{.03 in}3},...,\text{selected}_{\hspace{.04 in}m}\hspace{-0.03 in}\}\rangle \; \mapsto$ prefixfree$\left(\hspace{-0.02 in}s_{\text{selected}_1}\hspace{-0.03 in}\right)$ || prefixfree$\left(\hspace{-0.02 ...


3

I've been toying around with your function, and I've come to the conclusion it's not memory hard. The amount of required memory can be reduced to at maximum digestsize * 3 * rounds. The first problem is that the entropy does not avalanche throughout the state, but stays localized. For example, after 1 round the state of the 2nd block only depends on the ...



Top 50 recent answers are included