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24

As Chris Smith notes in the comments, HMAC is a specific MAC algorithm (or, rather, a method for constructing a MAC algorithm out of a cryptographic hash function). Thus, HMAC can be used for any application that requires a MAC algorithm. One possible reason for requiring HMAC specifically, as opposed to just a generic MAC algorithm, is that the HMAC ...


18

This is something I tend to disagree somewhat with Colin Percival on. You should use Encrypt-then-HMAC if and only if you can get it right. The biggest pitfall is using a short-circuiting string comparison versus a constant-time string comparison. Given the former, people can use timing attacks to forge valid HMACs for arbitrary ciphertexts. With an ...


18

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 ...


17

In short: You must authenticate the IV. Which particular attacks apply if you don't depends on the block cipher mode; I will give two common examples. In CTR mode, an attacker who fiddles with the IV can forge authenticated messages, but the content of the corresponding plaintext is beyond his control (since he doesn't know the key). Depending on the ...


16

You're missing the most important strength of HMAC: it comes with a proof of security (under some plausible assumptions). The outer key plays an important role in the proofs. The best place to learn more is to read the HMAC papers: Message authentication using hash functions: The HMAC construction, Mihir Bellare, Ran Canetti, Hugo Kawczyk, CryptoBytes ...


16

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 ...


15

Brute forcing the key would hardly be an issue: 128-bit keys (assuming they have been properly generated) are in a space which is way too large to be successfully explored by brute force; and 256-bit keys (the kind you put in AES-256) are even more larger. Whether AES is "faster" than HMAC or not does not make such brute force more feasible: even if each key ...


15

Actually, HMAC might still be secure for a hash function that is broken (with respect to the requirements of a cryptographic hash function, such as primary preimage resistance, secondary preimage resistance and collision resistance), but it must not be too badly broken. If you read the original paper, you see that the authors assume things such as the hash ...


15

Those "magic numbers" are related to the security proof behind the HMAC construction. In their Crypto'96 paper, Bellare, Canetti and Krawczyk first prove that $\mathrm{NMAC}_{(k_1, k_2)}(x) = F_{k_2}(F_{k_1}(x))$ forms a secure MAC ("message authentication code") provided $F_k(\cdot)$ is an iterated and keyed compression function enjoying some good ...


15

Yes, there are currently no known attacks on HMAC-MD5. In particular, after the first collision attacks on MD5, Mihir Bellare (one of the inventors of HMAC) came up with a new security proof for HMAC that doesn't require collision resistance: "Abstract: HMAC was proved by Bellare, Canetti and Krawczyk (1996) to be a PRF assuming that (1) the underlying ...


13

The short answer is: 2128 operations, no known birthday-like attack. The long answer: when HMAC was first published, it came with a security proof, tailored for iterated constructions like Merkle-Damgård. In a MD hash function (MD4, MD5 and the whole SHA family are MD hash functions), the data to hash is processed by blocks with a compression function: the ...


13

A Message Authentication Code (MAC) is a string of bits that is sent alongside a message. The MAC depends on the message itself and a secret key. No one should be able to compute a MAC without knowing the key. This allows two people who share a secret key to send messages to each without fear that someone else will tamper with the messages. (At least, if ...


11

The original security proof of HMAC, as well as a new one not requiring collision-resistance of hash, are for the construction hash(o_key_pad ∥ hash(i_key_pad ∥ message)) with o_key_pad different from i_key_pad (and both filling a block). That's the rationale for at least one of the constant. The other plays no role, it just must be different from the first. ...


11

HMAC was there first (the RFC 2104 is from 1997, while CMAC is from 2006), which is reason enough to explain its primacy. If you use HMAC, you will more easily find test vectors and implementations against which to test, and with which to interoperate, which again explains continued primacy. Being the de facto standard is a very strong position. On many ...


8

I really don't have an answer (other than saying that storing a hash of the password is good as any other way of solving your immediate problem; there are other ways, but they all allow an attacker to run a dictionary attack on the database). On the other hand, I do have these comments on what you're doing: If getting decrypted gibberish will really crash ...


8

No, you are not leaking any information except how to MAC those specific values with the specific key you are using. Using a short message is exactly as secure as using a long message. For the following, remember the definition HMAC (K,m) = H((K ⊕ opad) || H((K ⊕ ipad) || m)). There are two hashes here, an outer hash and an inner hash nested inside the ...


8

Yes, this would be secure. CTR (Counter) mode based on keyed function $F_K$ is secure as long as its output $$ W_i = F_K(i) $$ is unpredictable given previous outputs $$ F_K(1),F_K(2),\ldots,F_K(i-1). $$ This requirement is essentially the definition of a pseudo-random function (PRF). Most HMAC instantiations with widely used hash functions are believed to ...


8

The Keccak submission says: From the security claim in [12], a PRF constructed using HMAC shall resist a distinguishing attack that requires much fewer than $2^{c/2}$ queries and significantly less computation than a pre-image attack. Here, $c$ denotes the capacity of the sponge, i.e. the effective size of the internal state in bits. Since HMAC is a ...


8

It looks like, given your adversary model, things should be secure. HMAC as a randomness extractor has been shown to be good, especially when we can assume the hash function is collision resistant. That paper also has some results which tell how you could guard against the collision resistance being broken (basically use a hash function with larger output ...


7

Clearly, if you had been using AES-256-CBC for confidentiality and AES-256-CBC-MAC for authentication, it would not be secure to use the same key for both confidentiality and authentication. Hence, using the same key for confidentiality and authentication cannot generally be secure; you need additional premises to arrive at that conclusion. In your case it ...


7

As a Skein co-author, one of the properties of the UBI chaining mode is to give you HMAC-like properties in one pass. Skein itself consists of the Threefish tweakable block cipher, the UBI chaining mode, and some proofs that extend tweakable block cipher theory into a tweakable hash function theory that reduces the security of the hash function to the ...


7

Comparing a brute force attack on DES (with $2^{56}$ operations) to a birthday attack on CMAC (with $2^{64}$ operations) would appear to be an apples-to-Volkswagen comparison; they are assuming two things are similar, when they really aren't. The brute force attack on DES involves obtaining a single plaintext/ciphertext block pair, and then going through ...


7

TL;DR, an HMAC is a keyed hash of data. A good cryptographic hash function provides one important property: collision resistance. It should be impractical to find two messages that result in the same digest. An HMAC also provides collision resistance. But it also provides unforgeability. In order to generate an HMAC, one requires a key. If you only share ...


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

Is the calculated MAC encrypted using AES? What is the purpose? How about signing and verifying? How does AEs Play a role here? Is the case here that the encrypted AES is HMACed for signing and the HMAC is verified No, the MAC is not encrypted per se, however, it is calculated in conjunction with a key (independent of the encryption key). Simply ...


6

The usual ways to check that a user-supplied encryption key is correct are to either: store a (salted) hash of the key, and check that it matches, or encrypt a (partially) known block of data with the key and check that the decrypted output has the expected form. The former method is exactly same as what your OS, for example, does to verify that you ...


6

It is indeed safe to send it along with the ciphertext; the attacker can't learn anything from it (other than possibly how many packets has been generated so far, if you use a counter to generate the IVs), and if the attacker modifies the IV, the resulting message will fail to decrypt (with high probability). Existing protocols that can use GCM (TLS, IPSec) ...


6

CodeInChaos has it right about the infeasbility of this against a random key; however, lets run the numbers to see how extremely correct he is: Let us assume we are attacking HMAC-MD5 within TLS; this has a 128 bit key. The fastest GPU server (actually, it has 25 GPUs internally) can test about 400 billion keys per second. Let us assume that we, having a ...


6

This is highly insecure, for the same reason that ECB mode and simple substitution ciphers are. Every time you use the word the in your message, it will be encrypted the same way. The same goes for other, lower-frequency (but still fairly common) words -- like as or with or will (or any of hundreds of other examples). This is a humongous clue to ...


6

Password hashes need first pre-image resistance and should not cause many collisions among typical passwords (preserve the entropy). This collision "attack" violates neither requirement and causes no practical security issues. While this issue can find trivial collisions, they're not between commonly chosen passwords. A SHA-1 hash (and thus the shorter of ...



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