Tag Info

Hot answers tagged

19

I sent an email to Ron Rivest and got an answer back. The digits of $\pi$ are used as a sort of random number generator that is used in the Durstenfeld shuffle (see also Knuth vol 3, sec 3.4.2). Below is some pseudocode adapted from the description and code he sent me. S = [0, 1, ..., 255] digits_Pi = [3, 1, 4, 1, 5, 9, ...] # the digits of pi def ...


15

I do worry, but not for the resistance of SHA-3; I worry for its acceptance. Technically, what NIST wants to do is sound. They do want to somehow "break" a traditional rule, which is that a hash function with an output of n bits ought to resist collisions with strength 2n/2, and preimages (first and second) with strength 2n. Instead, NIST wants harmonized ...


15

The most efficient related-key attacks on AES-256 and resulting weaknesses AES-256-based hash functions are summarized in my PhD thesis. Though collision and preimage attacks on hash functions are out of reach yet, the components of these functions still expose some properties that are not expected of good hash functions or random oracles. Getting to the ...


15

Length extension attack The reason why $H(k || m)$ is insecure with most older hashes is that they use the Merkle–Damgård construction which suffers from length extensions. When length extensions are available it's possible to compute $H(k || m || m^\prime)$ knowing only $H(k || m)$ but not $k$. This violates the security requirements of a MAC. Like all ...


13

Most hashes are built from permutations (either keyed permutations/block-ciphers, as in MD5, SHA-1 and SHA-2, or unkeyed permutations as in Keccak/SHA-3 and CubeHash). A permutation is a shuffling of the inputs. Once you have a good random permutation, you can easily build a hash from it. See Construction of One-way compression functions from block ciphers ...


13

$Encrypt(m|H(m))$ is not an operating mode providing authentication; forgeries are possible in some very real scenarios. Depending on the encryption used, that can be assuming only known plaintext. Here is a simple example with $Encrypt$ a stream cipher, including any block cipher in CTR or OFB mode. Mallory wants to sign some message $m$ of his choice. ...


12

Unless Keccak has structural weaknesses that I am not aware of, the answer is surprisingly neither 128 nor 256! Gilles Brassard, Peter Høyer and Alain Tapp describe a sort of quantum birthday attack in their paper "Quantum Cryptanalysis of Hash and Claw-Free Functions" that effectively works by creating a table of size $\sqrt[3]{2^b}$ (versus the ...


11

MD5 and SHA-1 have a lot in common; SHA-1 was clearly inspired on either MD5 or MD4, or both (SHA-1 is a patched version of SHA-0, which was published in 1993, while MD5 was described as a RFC in 1992). The main structural differences are the following: SHA-1 has a larger state: 160 bits vs 128 bits. SHA-1 has more rounds: 80 vs 64. SHA-1 rounds have an ...


11

There does appear to be some confusion with point 1. The confusion probably stems from the fact that Keccak has an output size number and a capacity. Output size has little to no effect on security strength. Capacity is what really determines the security strength. So when the post says NIST will only standardize two security levels it is correct (as far as ...


11

The answer rather depends on what you mean by 'entropy'; if you mean 'Shannon Entropy', then no, a deterministic function cannot increase entropy. For example, if the unhashed password has only 7 different possible values, then the hashed version of the password will also have (at most) 7 different possible values; you've made things look more obscure, but ...


9

In addition to the performance problems poncho already mentioned when using RSA signatures without hashing I just want to add on the security warning of poncho: Reordering If you have a message $m>N$ with $N$ being the RSA modulus, then you have to perform at least 2 RSA signatures as $m$ does not longer fit into $Z_N$. Let us assume that it requires ...


9

For an adversary not knowing the definition of SHA-512 (or just not knowing the 512-bit initialization constant of SHA-512, defined as the first sixty-four bits of the fractional parts of the square roots of the first eight prime numbers), the sequence obtained by $$\begin{align*} H_0&=\text{SHA-512}(Seed){\small\text{ where }}Seed{\small\text{ is the ...


8

What Richie Frame describes above is correct. This is how most FDE solutions work. A new random encryption key is created whenever new container is created or disk is encrypted. That encryption key (often called Master Key) is then protected by users' password. In case of Truecrypt, master key is stored in volume header (link) and volume header is encrypted ...


8

I don't see any obvious security problems in your approach. You can look into key derivation functions, that can provide some additional security in case one of the following occurs: Your password leaks Your secret number leaks A weakness is identified in the hash function There is a few usability issues, that would have to be addressed as well: ...


8

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 ...


8

Shared secret resulting from the Diffie-Hellman step is a mathematical object; namely, the X coordinate of a curve point. It is a value in a non-binary range; moreover, it is indistinguishable from randomness only up to the security against discrete logarithm, i.e. about 128 bits. Thus, it is at least debatable that parts of the key might be guessable from ...


8

The generic model for a MAC is the following: the attacker is given access to a block box which implements the $S$ function with a key $k$ that the attacker does not know of. The attacker is allowed to make $q$ requests to the box on messages that he can choose arbitrarily. The goal of the attacker is to make a forgery, i.e. produce values $m$ and $t$ such ...


8

What Dan Boneh says is not a formal definition as you want it. Let me quote Rogaway on this: In cryptographic practice, a collision-resistant hash-function (also called a collision-free or collision-intractable hash-function) maps arbitrary-length strings to fixed-length ones; it’s an algorithm $H:\{0,1\}^*\rightarrow \{0,1\}^n$ for some fixed ...


8

The answer to your edited question is "yes, it is possible". As a trivial example, let $H$ be an ideal $k$-bit hash function. Due to the existence of the generic birthday attack, $H$ provides only about $k/2$ bits of collision resistance — that is, an attack can, on average, find a collision after about $2^{k/2}$ hash function evaluations. Denote ...


8

MD5 is ok here as usual cryptographic attacks do not apply in this scenario. The probability of accidental MD5 collision is much less than usual probability for soft error. For details read more. MD5 is currently considered too weak to work as a cryptographic hash. However, for all traditional (i.e. non-cryptographic) hash uses MD5 is often perfectly ...


8

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 ...


8

With the definitions that a function $F$ is collision-resistant when a [computationally bounded] adversary can't [with sizable odds] exhibit any $(a,b)$ with $a\ne b$ and $F(a)=F(b)$; first-preimage-resistant when, given $f$ determined as $F(a)$ for an unknown random $a$, a [computationally bounded] adversary can't [with sizable odds] exhibit any $b$ with ...


8

If you mean exactly as likely, no, because the number of possible hashes is not a multiple of $100$. This is assuming all the hashes are exactly equally likely. You can come very close just by taking $SHA256 hash \pmod {100}$ This will be within one part in $\frac {2^{256}}{100}$, which is a very small number. If you want truly equal, check that the hash ...


8

When only using one-way hashing, is it possible to tell the number of characters changed between the old and new password? No. If the hash function is strong, even a single bit change will give a completely different hash. The only way to tell how many characters differ between a particular unknown hash value and a known password would be an exhaustive ...


7

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) ...


7

Well, how resistant to attack would depend on what security properties you would need from it. There are three standard assumptions we can make about a hash function: Given a hash value, it is difficult to find an image that hashes to that value; this is known as preimage resistance Given a image that hashes to a specific value, it is difficult to find ...


7

Uniformity is a tricky one. SHA-256 (as well as SHA-3 for that matter) follows a heuristic approach. That is, the design is not based on a hardness assumption (for example, the factoring or discrete-log assumption) but on criteria that have only been verified empirically. As such, also the study of uniformity is an empirical study. The development of ...


7

They are actually the same, because you missed W^=V in the second link. When you work out the XORs, you arrive at the same constants.


6

There are some attacks on hashes keyed with a secret suffix. The proper primitive for deriving a secret from keys/passwords and an identifier is a key derivation function. In your case, if the secret number is random a fast key derivation function, like HKDF, would be enough to expand the key into several site-specific hashes. In that case there's no need ...



Only top voted, non community-wiki answers of a minimum length are eligible