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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

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. ...


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

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 ...


11

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 ...


11

I restrict to hash functions $H$ with an output of some fixed size $n\ge1$ bit(s), accepting as input some strings, including all $n$-bit strings; MD5 (resp. SHA-1, SHA-256) is an example of such function for $n=128$ (resp. $n=160$, $n=256$). Whether there exists a solution to $H(x)=x$ depends on the particular hash function. If $H$ is a random function (as ...


10

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 ...


10

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 ...


9

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: ...


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

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

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

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

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.


7

A "generic attack" against a cryptographical primitive is one that can be run independently of the details of how that cryptographical primitive is implemented. The most obvious case is a cipher that takes an $N$ bit key; the generic attack of brute force takes a ciphertext, and attempts to decrypt it with all $2^N$ keys; when we find the known (or ...


6

No such function with either property would meet the requirements of a secure hash function; either of those properties would make it easy to find preimages, that is, given a value $H(x)$, you can find a value $y$ with $H(y) = H(x)$. First off, I assume that $n$ is a constant for the hash function; if we were to assume that the first property holds for any ...


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 ...


6

PBKDF2 (as defined by RFC 2898) is a function of the form $$DK = \text{PBKDF2}(\text{PRF}, Password, Salt, c, dkLen)$$ In most practical use cases, the $\text{PRF}$ is $\text{HMAC}$ instantiated with a Merkle-Damgård hash function such as $\text{SHA-1}$. The time to compute $\text{PBKDF2}$ is roughly linear with the iteration parameter $c$, all other ...


6

Well, one reason to hash the data before signing it is because RSA can handle only so much data; we might want to sign messages longer than that. For example, suppose we are using a 2k RSA key; that means that the RSA operation can handle messages up to 2047 bits; or 255 bytes. We often want to sign messages longer than 255 bytes. By hashing the message ...


6

A hash function is a hash function: a deterministic, public function with no secret parameter. Everybody can compute SHA-256, and they all get the same results when applied on the same inputs. Therefore, if you give H(m) to the attacker (who tries to recover m: a preimage attack), giving him H(H(m)) too does not grant him any extra actual help -- because he ...


6

The resume of that other answer could be: When you have a password hashed, it's hard (very hard) to find out what was the original password: you have to try all combinations, until you find the hash. That's brute-force. Someone can speed up a bit this process, by pre-computing many passwords: he'll store all those passwords / hashes, and will try to find ...



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