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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 ...


9

What if text would replaced with H(text)? Will it weaken the HMAC algorithm? Yes. It makes collision attacks on the hash function apply to the MAC, which isn't normally the case with HMAC. You can find a pair $(m, m')$ that hash to the same value, get the MAC for one of them and move it to the other. That means that your modified HMAC construction ...


6

Summary: a single HMAC-MD5 with a key later revealed is a completely insecure way to do commitments of messages that can be chosen malignantly, because of the ease with which MD5 collisions can now be found. There is no compelling evidence that's so insecure for messages constrained to belong in a small arbitrary set that no adversary can choose or ...


6

Will it weaken the HMAC algorithm? No, assuming that $H$ is a collision resistant hash function (which is a stronger constraint than what HMAC puts on the hash function). The security property that a MAC (such as HMAC) has is that "given a large set of $message, tag=HMAC(K, message)$ pairs (where $K$ is an unknown key), where the attacker can choose ...


5

First of all there does exist information theoretically secure message authentication codes suitable for use with a one time pad. An HMAC is not one of those information theoretically secure. As far as I recall the first article presenting such a construction is the 1981 article by Wegman and Carter: New hash functions and their use in authentication and ...


5

NMAC is really just an "education tool" on the way to HMAC and I don't think anyone intended it to be used. The two keys are needed since the first and second hashes have different purposes. The first hash on the message is just needed to get collision resistance, whereas the second hash is supposed to provide a pseudorandom function type property. As such, ...


4

Actually you are quite near on implementing PBKDF2. It is kind of iterated HMAC execution. So have a look here and just implement the missing parts: PBKDF2


4

AES CBC usually requires padding, such as PKCS#7 padding. This padding is 1 to 16 bytes, 16 being the block size of AES. The HMAC will add 256 / 8 = 32 bytes to the total. Usually you will need to store the randomized IV as well with ciphertext, to allow for reuse of the key, adding another 16 bytes (the block size again). So the total overhead will be about ...


4

The Encrypt then MAC is done in general in order to be sure to decrypt into the correct plaintext, without risking of parsing a non-authentic plaintext message. If you don't MAC the IV, then Mallory (attacker that can tamper with messages as a man-in-the-middle) can modify the IV and your MAC will be still validated as good. So you will decrypt into an ...


4

The construction you are proposing is called the "envelope" or "sandwich" MAC, it predates HMAC, and it is in fact secure—provided the key and message are appropriately padded. That is, $$ \text{SHA256}(k \parallel m \parallel 1 \parallel 0^{b - 1 - (|m| \bmod b)} \parallel k) $$ is secure, as long as $k$ is the underlying hash function's block length $b$ ...


4

In summary: Yes, HMAC is the way to go for construction of a MAC from an arbitrary concrete iterated hash. We have no constructive argument of security of the MAC constructs in the question; we even have a concrete attack when using some otherwise apparently fine hashes. I consider a hash constructed by iterating a compression function $F$ as ...


3

Rejecting replays is the duty of a higher level protocol. Simple authenticated encryption will accept any message with a valid MAC, even if you receive it several times. Decryption is a stateless process, but you need state to keep track of messages you already received. For example you could associate an increasing counter for each message you send. The ...


3

This is vulnerable to a length extension attack. Given a valid nonce/MAC, the nonce can be extended to forge a new valid nonce/MAC value. This is because $m_4$ is appended to the end inside the outer hash. How this affects you will depend on how you validate your nonce. But in general, this is not a secure construction. There's probably more things wrong ...


3

I know SHAKE128 and 256 are part of the SHA-3 standard but is the SHA3 standard officially released yet? i can only find a draft of the publication, does this mean it's not official and therefor not proven to be secure? No, SHA-3 has not been formally approved. On the other hand, what do you mean "not proved to be secure"? Do you really thing that ...


3

First, terms: A MAC is a generic term for a class of cryptographic primitives. It's in the same category as "hash" or "PRNG." HMAC is a particular construction that, combined with a suitable cryptographic hash, gives a secure MAC function (it can also be used to generically refer to any HMAC algorithm, since HMAC is secure with pretty much any standard hash, ...


3

I'm going to agree with @fgrieu's marvelous post above in a back-handed way. My answer is: No, you don't have to use an HMAC. Do it anyway. As you noted, some hashes, sush as SHA-3 (especially in its Keccak form), Skein (which I was a team member on), and others will work just fine. In the case of Skein, there is a one-pass Skein-MAC that has a proof of ...


3

When people say HMAC-MD5 or HMAC-SHA1 are still secure, they mean that they're still secure as PRF and MAC. The key assumption here is that the key is unknown to the attacker. $$\mathrm{HMAC} = \mathrm{hash}(k_2 | \mathrm{hash}(k_1 | m)) $$ Potential attack 1: Find a universal collision, that's valid for many keys: Using HMAC the message doesn't get ...


3

In the first section of this answer I'll assume that through better hardware or algorithmic improvements, it has become routinely feasible to exhibit a collision for SHA-1 by a method similar to that of Xiaoyun Wang, Yiqun Lisa Yin, and Hongbo Yu's attack, or Marc Stevens's attack. Neither has ever been achieved publicly, but it is clearly feasible (the ...


3

If you are asking if $M(M(M(k, C_1) C_2), s)$ has the same or better security as $M(k, C_1 | C_2 | s)$ then the answer is yes. You can see this as a double key derivation to calculate a new key from $k$ using the constants $C_1$ and $C_2$ as derivation data. Then the resulting key is used to MAC the final serialNumber. $M$ is of course HMAC, $C_1$ and $C_2$ ...


2

Actually the HMAC value is not decrypted at all. The recipient takes all the needed input and she computes the HMAC on her own side and check if the result she got it is equal to the value on the message she got. You can roughly see the HMAC algorithm as an symmetric key signature. You cannot decrypt an HMAC, you only check that the value is correct.


2

PBKDF2 is defined for an arbitrary PRF, but in practice HMAC is usually used. Either with SHA-1 (original definition), SHA-256 (e.g. in scrypt) or even SHA-512-256 (NaCL). So first you can look at how swapping key and message affects HMAC. HMAC has two cases depending on the length of the key: if it's no longer than the block size of the hash, it is used as ...


2

Yes, you are reading this right. The requests for random value from NIST 800-90 drbgs perturbed the state. If this is a problem you can add a layer that optionally buffers values and always makes constant size requests.


2

BCrypt is considered more secure The theoretical security of bcrypt has received less scrutiny than that of PBKDF2, SHA2 and HMAC. PBKDF2 is thus widely standardised (e.g. in NIST SP800-132 and PKCS #5) while bcrypt is not. In practice the security (resistance to brute force attack or dictionary attack) of bcrypt and PBKDF2-HMAC-SHA512 can be ...


2

I understand the system as follows: data blocks are enciphered per AES-CTR, using key encryption_key, with an IV made by concatenating device_id and a counter held in Flash or EEPROM, incremented at each use; that enciphered data is integrity-protected by a 256-bit mac_tag computed using HMAC-SHA256 and mac_key. That's theoretically sound if device_id ...


2

There is no known inherent weakness in HMAC-SHA256 that requires key rotation for a suitably secret key. However, other aspects of the system in which you are using HMAC-SHA256 may make key rotation necessary. For instance if you don't want the same message to have the same hmac over time for some reason a key rotation would achieve that (but so would a ...


2

Fgrieu has already posted a good answer, which I won't try to repeat. However, here are a few additional observations: For an embedded system, you may want to consider using CMAC-AES instead of HMAC, since you can reuse your AES implementation, and don't need a separate hash function. Further consider using SIV mode (RFC 5297). It's very similar to ...


2

That's a lot of questions, I'll try and answer in order. A hash or message digest alone is not secure because anybody can calculate and thus substitute a hash value. If you (correctly) add a key to the mix then you get a HMAC, which can be used. Nowadays often a HMAC is used, or an authenticated mode of authentication such as GCM, CCM (for packet ...


2

Well, there are two potential key recovery attacks against HMAC (assuming a reasonable hash function): Brute force the key; that is, take a valid (Message, MAC) pair, and try every possible key, and look for a key that gives that MAC for that Message Brute force the internal hashing state immediately after processing the IPAD/OPAD; here, you would take a ...


2

If the keys and messages are known, yes, you can distinguish which were used - because you can test them all. If not, then this is "sligtly harder" (= not really possible with big enough values). Anything of the further answer will assume that the attacker doesn't know the keys or the messages. The definition of HMAC looks like this: $HMAC(K, m) = H(K ...


2

Not, that should be not possible, at least if you have a good key. All this is only valid as long as SHA-256 is still a secure hash algorithm and not broken. To be precise, as long as HMAC-SHA-256 is not broken. Does the attacker have any informations about the messages (other than the length and that they are "random-noise")? If he/she doesn't: No, ...



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