# Tag Info

4

If you add a truly random character into a truly random position of a word (uniformly chosen), you get "entropy of position" + "entropy of character" as addition to the entropy of the word. (Not exactly, it's a bit less). The entropy of character is the size of the possible characters. 64 possible characters would be $log2(64) = 6$ bits of entropy. Entropy ...

3

Solving a 256-bit discrete log is absolutely doable, and quite quickly, these days; there are public tools that can do it, though they may require some expertise to use. On that note, even a 1024-bit modulus is not particularly conservative: it is generally agreed that well-funded organizations today could break logs of that size as well, but at a very ...

2

Being able to solve the discrete logarithm in SRP-6 allows an eavesdropping attacker to dictionary attack the password. It will not directly reveal a strong password or its hash. It requires the attacker to observe a successful authentication, $B$ alone does not suffice. The attacker eavesdrops $s$, $A = g^a$, $B$ and $M_1$. The attacker solves $a$ from ...

2

"Would it be possible for an attacker to launch an offline dictionary/brute-force attack on the B public key: ..." That is possible if and only if the attacker can distinguish b's distribution from the uniform distribution on {0,1,2,3,...,N-3,N-2}. $\:$ If so, an attacker could compute verifiers v for candidate passwords, subtract kv from B mod N, and ...

2

The question is well answered by ninefingers, but the question exposes fundamental confusion on the part of Mitchell. The NT Hash is not salted, and it IS PASSWORD EQUIVALENT. There is no need to use the hash to get the password, just use the hash to access the resource! Second, the password must be greater than 14 characters to avoid the LM hash, see MS ...

1

You're using bytewise xor, and three plaintexts $x,y,z$ with a small alphabet (A-Z, which I assume means ASCII values 0x41-0x5a). Then knowing $a \oplus x$, at a certain index, we know all possible values of $a$, at that same index, namely $\{0x41 \oplus (a \oplus x),\ldots, 0x5a \oplus (a \oplus x)\}$. But we have 2 more constraints from $a \oplus y$ and \$a ...

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