# Tag Info

37

SHA-512 truncated to 256 bits is as safe as SHA-256 as far as we know. The NIST did basically that with SHA-512/256 introduced March 2012 in FIPS 180-4 (because it is faster than SHA-256 when implemented in software on many 64-bit CPUs). SHA-224 is just as safe as using 224 bits of SHA-256, because that's basically how SHA-224 is constructed. What bits are ...

25

It would be very freakish if it turned out to be true. It is not an expected property of SHA-512 to have such bijectivity. It would be worrisome, even, because that's a kind of structure that should not appear in a proper cryptographic hash function. Actually proving that SHA-512, for 512-bit blocks, is not bijective, would already be a kind of a problem. ...

25

Would you use HMAC-SHA1 or HMAC-SHA256 for message authentication? Yes. That is a semi-serious answer; both are very good choices, assuming, of course, that a Message Authentication Code is the appropriate solution (that is, both sides share a secret key), and you don't need extreme speed. How much HMAC-SHA256 is slower than HMAC-SHA1? Those sorts ...

24

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 ...

17

The following table should provide a nice comparison of the SHA algorithms and their current status: [38] The theoretical attack on SHA-1 refers to “Freestart collision for full SHA-1” (PDF) by Marc Stevens and Pierre Karpman and Thomas Peyrin, first published 8 October 2015.

15

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 ...

12

Well, as far as we know, the mode you suggest should be secure. Now, to be honest, AES256 versus your mode isn't quite a fair comparison; your mode gives somewhat less theoretical security; if you encrypt a known $2^n$ block message, the key can be recovered with $2^{256-n}$ effort; however, this observation doesn't really affect the practical security. ...

11

If you want to use Skein (one of the SHA-3 candidates) anyway: it has a "mode of operation" (configuration variant) for tree hashing, which works just like your method 2. It does this internally of the operation, as multiple calls of UBI on the individual blocks. This is described in section 3.5.6 of the Skein specification paper (version 1.3). You will ...

11

Actually a tree-based hashing as you describe it (your method 2) somewhat lowers resistance to second preimages. For a hash function with a $n$-bit output, we expect resistance to: collisions up to $2^{n/2}$ effort, second preimages up to $2^{n/2}$, preimages up to $2^n$. "Effort" is here measured in number of invocations of the hash function on a short,...

11

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: $$I_0 = \left \lfloor \mathrm{frac} \left (\sqrt{23} \right ) · 2^{64} \right \rfloor$$ $$I_1 = \left \lfloor \mathrm{frac} \left (\sqrt{29} \right ) · 2^{64} \right \rfloor$$ $$\cdots$$ ...

11

In early years of hash function design it was unclear how to choose constants (not only initial vectors), and it was widely assumed that the more random they look, the more secure the function is. There is still not much research in this direction. However, there have been several attacks (rotational cryptanalysis, slide attacks, internal difference attacks) ...

11

SHA-224, part of FIPS 180 since FIPS 180-3 FIPS 180-2 change notice 1 of 2004, was introduced to match the second of the security strengths {80, 112, 128, 192, 256} defined in the document that became NIST Special Publication 800-57 – Recommendation for Key Management – Part 1: General (Revision 3). That security strength itself was kept ...

11

The functions used by SHA-2, called $Ch$ and $Maj$ are defined like this in the standard: $$Ch(x, y, z) = (x \land y) \oplus (\lnot x \land z)$$ $$Maj(x, y, z) = (x \land y) \oplus (x \land z) \oplus (y \land z)$$ However, an equivalent way to define them replaces the XOR with OR, as the standard (pdf) states: Each of the algorithms include Ch(x, y, ... 10 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 ... 10 I would use HMAC-SHA256. While poncho's answer that both are secure is reasonable, there are several reasons I would prefer to use SHA-256 as the hash: Attacks only get better. SHA-1 collision resistance is already broken, so it's not impossible that other attacks will also be possible in the future. It allows you to depend on just one hash function, ... 9 Your cipher looks a bit like the output feedback mode of operation for block ciphers. While OFB for block ciphers is considered safe (as long as it is used right), OFB for a hash function like you are using it has the problem that the key is only used at the start, to generate the "initialization vector", not at each step of the algorithm. Thus, as ... 9 First of all, this no block-cipher at all. It's a stream-cipher. Thus you can use every key only once, and you can't use any cipher modes built on block-ciphers. Your scheme is vulnerable to a known plaintext attack. If the attacker knows 32 aligned(or 63 unaligned) bytes of plaintext, he can calculate the state of your cipher: S_i = P_i \oplus C_i $... 9 Adding more qubits does not increase the computation speed. A quantum computer with 4 qubits does not factorize faster than one with 2. The qubits are the "memory" of the quantum computer. More qubits mean you can factor bigger numbers. If I remember correctly, you need a superposition of$\Theta(N^2)$terms, which means$\Theta(\log(N^2))$qubits to factor ... 9 The attack which you link to, on ECDSA, is related to the following: the signer computes several values$kG$, for random$k$values chosen uniformly modulo$n$($n$is the size of the subgroup generated by$G$). One such value is generated for each signature. It is important that the selection is uniform: even small biases can be exploited in order to make a ... 9 No, because even SHA-512 was considered overkill from a security perspective. It has 256-bit collision resistance, which is unbreakable. (The link is about keys but a similar argument applies.) If you think large quantum computers will be efficient, a 512-bit hash makes some sense, but even then a 1024-bit one wouldn't. A quantum computer requires$O(2^{n/3}...

8

No. Cryptographic hash functions model a random function, not a random permutation. A significant fraction of output hash values are expected to be unreachable and another fraction have multiple preimages. While bijectivity in general does not mean that the inverse is easy to calculate, for the types of constructs which are used in hash functions in ...

8

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 ...

7

SHA-1, SHA-224 and SHA-256 append the bit “1” to the end of the message, followed by k zero bits, where k is the smallest, non-negative solution to the equation l+1+k ≡ 448 mod 512, where l - message length. In second step they use 32-bit words. SHA-384, SHA-512, SHA-512/224 and SHA-512/256 use different equation: l+1+k ≡ 896 mod 1024 and in 2. step ...

7

Well, SHA-1 and SHA-256 are both limited to inputs of no more than $2^{64}-1$ bits; the HMAC architecture itself prepends a logical IPAD (which is 512 bits); hence both HMAC-SHA160 and HMAC-SHA256 are both limited to inputs of no more than $2^{64} - 513$ bits, which is about 2 exabytes. I rather suspect that this is not a serious limitation to your ...

7

The main thing that makes HMAC secure in typical use even with MD5 is that it is used with a secret key. That means only preimage attacks are really relevant, since finding a collision is always an online attack if you don't know the key. With known attacks the preimage resistance of both MD5 and SHA-1 is > 100 bits. Additionally, HMAC may be secure even ...

6

If the hash function is any good, then it should behave as a "random function" (i.e. a function chosen randomly and uniformly among all possible functions). For a random function with output size $n$ bits, it is expected that nested application will follow a "rho" pattern: the sequence of successive values ultimately enters a cycle with an expected size of $... 6 Revised: The proposed construction is just fine, and in particular: at least as secure as SHA-256 against collision attacks, that is the ability for an adversary to construct two files with the same hash; likely about as secure as SHA-256 against both first and second preimage attacks, that is the ability for an adversary to construct (for first preimage) ... 6 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 ...

6

First, md5, Can I use md5 as a two way function if I can break the data in 64bit, will I be able to recover the original message without a pre-calculated a MD5 is a hash function, not a cipher. Differently stated: you will not be able to encrypt or decrypt anything by simply using a hash function. You could compare MD5 hashes with each other, but that ...

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