# Tag Info

1

You can do this slightly better with an additional $\mathcal{0}(2^{56})$ memory and with $\mathcal{0}(2^{56})$ time. You can notice that the relation $c \leftarrow E_{k_1}(E_{k_2}(m))$ can be rewritten as $D_{k_1}(c) = E_{k_2}(m)$ (just apply the decrypt function on both sides. First step consists in the generation of every pair $(k_2, E_{k_2}(m))$ and ...

5

As Gerald Davis explained in the other answer, there are about 6 million possible passwords, which is way too few. However, there's an additional weakness: since the password and salt are combined with XOR rather than concatenation, it is sufficient to generate a table for all hash values. If you know the $x$ for which $H(x) = h$, you know that the password ...

6

Possible password search space = $36^5$ = 6.05 million possible combinations or ~$2^{26}$. If the passwords were randomly generated it would be 26 bits of entropy which isn't just weak it is pointless. To put that into perspective the throughput on modern GPUs is on the order 1 billion SHA-256 hashes per second. So the exhaustive search time to break an ...

Top 50 recent answers are included