# Tag Info

82

@Ninefingers answers the question quite well; I just want to add a few details. Encrypt-then-MAC is the mode which is recommended by most researchers. Mostly, it makes it easier to prove the security of the encryption part (because thanks to the MAC, a decryption engine cannot be fed with invalid ciphertexts; this yields automatic protection against chosen ...

64

These types of cryptographic primitive can be distinguished by the security goals they fulfill (in the simple protocol of "appending to a message"): Integrity: Can the recipient be confident that the message has not been accidentally modified? Authentication: Can the recipient be confident that the message originates from the sender? Non-repudiation: If ...

27

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 ...

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Hugo Krawczyk has a paper titled The Order of Encryption and Authentication for Protecting Communications (or: How Secure Is SSL?). It identifies 3 types of combining authentication (MAC) with encryption: Encrypt then Authenticate (EtA) used in IPsec; Authenticate then Encrypt (AtE) used in SSL; Encrypt and Authenticate (E&A) used in SSH. It proves ...

24

The word "secure hash function" usually means (for a function $H$) Preimage resistance: Given a value $h$, it is hard to find a message $x$ so that $h = H(x)$. Second preimage resistance: Given a message $x$, it is hard to find a message $x' \neq x$ such that $H(x) = H(x')$. Collision resistance: It is hard to find two messages $x$, $x'$ such that $H(x) = ... 21 A Message Authentication Code (MAC) is a string of bits that is sent alongside a message. The MAC depends on the message itself and a secret key. No one should be able to compute a MAC without knowing the key. This allows two people who share a secret key to send messages to each without fear that someone else will tamper with the messages. (At least, if ... 20 Length extension attack The reason why$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$. This violates the security requirements of a MAC. Like all ... 19 The nCipher Advisory #13 cited in your securityfocus.com link contains the explanation of the vulnerability (in the section "Cryptographic details"). The CBC-MAC algorithm works similar to the CBC encryption algorithm, but only outputting the final block (or a part of this). Each block of the plain text is XOR-ed with the previous ciphertext and then ... 18$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$of his choice. ... 17 You're missing the most important strength of HMAC: it comes with a proof of security (under some plausible assumptions). The outer key plays an important role in the proofs. The best place to learn more is to read the HMAC papers: Message authentication using hash functions: The HMAC construction, Mihir Bellare, Ran Canetti, Hugo Kawczyk, CryptoBytes ... 16 TL;DR No, the approach is not secure. Use a standard like CMAC instead. Or even better, check your AES accelerator module to see if it supports any AEAD modes of encryption like GCM, CCM, EAX. Long Version In order for a message authentication code (MAC) to be secure, an adversary with oracle access to the MAC (basically this means the adversary can send ... 14 The reason$H(k|m)$(where$|$is concatenation) is not the standard comes from the message extension attack. If I, as an attacker, have$H(k|m)$and$m$, I can compute$H(k|m|p|m')$(where$p$is the padding that$H$would have applied to$k|m$in computing the digest, and$m'$is an arbitrary message) without knowing$k$. I would then send ... 14 Although there are already many answers here, I wanted to strongly advocate AGAINST MAC-then-encrypt. I fully agree with Thomas' first half of the answer, but completely disagree with the second half. The ciphertext is the ENTIRE ciphertext (including IV etc.), and this is what must be MACed. This is granted. However, if you MAC-then-encrypt in the ... 14 If this requires a single answer among 1/2/3/4 (rather than none), I would select 3, by the following reasoning: Digital Signature provides confidentiality while message authentication code can not We can summarily exclude this, since since Digital Signature simply do not provide confidentiality. Digital Signatures works faster than ... 13 The MAC value should be calculated over all of the input, not just the first block. The chaining of CBC makes sure that the bits in the last block of ciphertext depends on all the previous blocks. 12 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 ... 11 In TLS (that's the standard name for SSL; TLS 1.2 is like "SSL version 3.3"), client and server ends up with a shared secret (the "master secret", a 48-byte sequence; when using RSA key exchange, the master secret is derived from the "premaster secret" which is the 48-byte string that the client encrypts with the server public key). That shared secret is ... 11 UMAC is described in full details in RFC 4418. When the RFC talks about "secret selection", it really means "there is a secret key involved here". UMAC works with universal hashing, which can be viewed as a family of hash functions, and a key which selects which hash function we are talking of. The term "hash function" might be a bit confusing here, because ... 10 No, in general, this is not secure, unless you make additional assumptions on the encryption method beyond the standard assumption of privacy. To simplify things a bit, the assumption of privacy means that given a ciphertext$C$, the attacker has no information about what the plaintext might be. However, in your case, we don't really care if the attacker ... 10 This construction is not secure. It was proposed in this paper in a quick sentence for possibly fixing the insecure secret prefix construction from the other question:$\mathcal{H}(k||m)$. The author then proposes and analyzes an enveloping method:$\mathcal{H}(k_1||x||k_2)$. An attack involving finding an internal collision applies to ... 10 Yes, this would be secure. CTR (Counter) mode based on keyed function$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 ... 10 First the theoretical explanations: Integrity and authenticity are different goals to achieve, but both are achieved (for symmetric encryption) with a MAC. You should probably be using encrypt-than-MAC or an authenticated cipher unless you have very good reasons not to. No blanket statements can be made though. HMAC: HMAC is a often used construct. It ... 9 I'll assumme All ciphered blocks means the same as ciphertext for CBC-Encryption with implicit zero IV, while CBC-MAC is the last block of that. All ciphered blocks is unsafe as a message authenticator for messages longer than one block, for it succumbs to a trivial attack (here with two blocks): Eve intercepts message$M=M_0||M_1$and its authenticator ... 9 In general, a MAC with a known fixed "key" is not a secure hash. That is, you can have a secure MAC (that is, someone without the key, but with a large number of message/MAC pairs, cannot come up with another valid message/MAC pair) that is not collision resistant, or even preimage resistant, if the attacker does know the key. In addition, you don't have ... 9 Moxie Marlinspike calls it in his article http://www.thoughtcrime.org/blog/the-cryptographic-doom-principle/ the doom principle: if you have to perform any cryptographic operation before verifying the MAC on a message you’ve received, it will somehow inevitably lead to doom. He also demonstrates two attacks which are possible because of trying to ... 9 It is not secure, because an attacker can "mix and match" the output blocks from different authentication tags on different input messages, or repeat output blocks for repeated input blocks. For example, if the attacker knows the tag$F_k(m)$for a one-block message$m$, then it can forge the correct tag$F_k(m) \mid F_k(m)$for the two-block message$m ...

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One rationale for avoiding randomized schemes in general, and in MACs in particular, is that the random in such schemes tends to increases the size of cryptograms or reduce the size of the payload. An example is scheme 2 in ISO/IEC 9796-2 RSA signature with message recovery, where the size of the random/salt field is directly antagonist with the amount of ...

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This scheme is not worth the name MAC; it is horribly weak. First and foremost, the tag/MAC is unchanged when two blocks of plaintext are exchanged (because of the commutativity and associativity of the $\oplus$ operation). If follows that from any message with at least two different blocks, we can make a different message for which we know the tag/MAC. ...

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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 ...

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