# Tag Info

11

No. The wikipedia article is in my honest opinion misrepresenting this article on a reduced round attack on the SHA-2 family of hashes. Although these attacks improve upon the existing reduced round SHA-2 attacks, they do not threaten the security of the full SHA-2 family. In other words, no collisions have been found in any of the SHA-2 hashes. The ...

6

The initial hash values for SHA-512 are the 64-bit binary expansion of the fractional part of the square root of the 9th through 16th primes (23, 29, 31, ..., 53). That is: $$I_0 = \left \lfloor \mathrm{frac} \left (\sqrt{23} \right ) · 2^{64} \right \rfloor$$ $$I_1 = \left \lfloor \mathrm{frac} \left (\sqrt{29} \right ) · 2^{64} \right \rfloor$$ $$\cdots$$ ...

5

With the message padding scheme of SHA-2/SHA-256 as it stands (add one 1 bit, a minimal number of 0 bits so that the overall padded message will end on a block boundary, then the original message length over some fixed number of bits), I know no attack enabled by allowing a different IV. However, allowing an arbitrary IV renders ineffective one of the two ...

5

The definitions given in FIPS 180-4 are $$\mathtt{Maj}(x, y, z)=(x∧y)⊕(x∧z)⊕(y∧z)$$ $$\mathtt{Ch}(x,y,z)=(x∧y)⊕(¬x∧z)$$ where $∧$ is bitwise AND, $⊕$ is bitwise exclusive-OR, and $¬$ is bitwise negation. The functions are defined for bit vectors (of 32 bits in case fo SHA-256). I'm positive $\mathtt{Maj}$ stands for majority: the result is set according to ...

4

The CTR mode of encryption is defined in general for any cryptographically strong pseudo-random function (PRF). You can build such a PRF from a hash function. For CTR, you produce a key stream by concatenating: $$F(k,0) || F(k,1) || ... || F(k,m)$$ where $F$ is your secure PRF, $k$ is your key, and $m$ is the the length of your plaintext divided by the ...

4

In short, no. So, what is the current state of cryptanalysis with SHA-1 (for reference only as this question relates to SHA-2) and SHA-2? Bruce Schneier has declared SHA-1 broken. That is because researchers found a way to break full SHA-1 in $2^{69}$ operations. Much less than the $2^{80}$ operations it should take to find a collision due to the birthday ...

2

There are 94 printable ascii characters. Not all of which are valid for file names, however. There should be $64=2^6$ that are valid for file names, so read $6$ bits at a time and map those to one of the $64$ characters that are valid for file names. That would give you $256/6\approx 43$ characters. It will be hard to get much smaller than that. That ...

2

What you explain in the question resembles SHACAL-2 cipher's forward cipher function, see http://en.wikipedia.org/wiki/SHACAL#Security_of_SHACAL-2. SHACAL-2 is NESSIE accepted way of using SHA-256 as cipher, therefore it has appeared somewhat secure.

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