Tag Info

New answers tagged

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

It seems that you are trying to implement your own KBKDF (Key Based Key Derivation Function) using HMAC. Maybe it is better to use a pre-defined one. It would be more sensible maybe to use an HSM that is FIPS certified for NIST SP 800-108. These use one of the KBKDFs defined in NIST SP 800-108. You can still use the idea of the random by putting it in the ...


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 ...


1

Here are some advantages and disadvantages for each of the three classes of MACs, which I know about: Based on block cipher There are constructions where the security of the MAC is proven in terms of the security definition of a block cipher. This means as long as the block cipher is secure, the MAC will be secure. There are constructions where encryption ...


1

From that same document: In any case the minimal recommended length for K is L bytes (as the hash output length). So as long as the key is fully randomized, i.e. a cryptographically strong key, the time for brute forcing is at least the same or higher than the time finding a collision. If your key is indeed smaller or not fully randomized, then brute ...


1

I do not recommend inventing your own protocol for this. Instead, I recommend using TLS client certificates. They solve exactly your problem, and solve it well. The TLS folks thought a lot about the security challenges involved in designing such a protocol, so you wouldn't have to. You mention that you perceive the complexity of client certificates to be ...


1

Unsurprisingly, any secure MACs are a secure choice. Assessing their relative security beyond how many bits of security they offer isn't possible in general. However, there are some differences that don't depend on the protocol: Unmodified CBC-MAC is only secure for fixed length messages, otherwise it allows some forgeries. Block cipher based MACs allow ...


5

As shown in the paper Ricky Demer linked in the comments, HMAC can be secure even when the underlying hash function is not collision resistant. Only PRF-ness of the hash function is required, and SHA-1 is not known to lack it. Or a couple of other conditions can suffice even if it isn't a PRF. Intuitively, it makes sense that HMAC is secure as a MAC even ...


3

A key derivation function lets you derive keys from others. In this case I would use HKDF, which means using HMAC in a predefined way. Your key material is the keys $X$ and $Y$, so you can concatenate those to get the PRK for HKDF-Expand. An output key would then be $\operatorname{HMAC}(X||Y, \text{info} || \text{0x01})$, if the size of the HMAC is long ...


1

HMAC is considered the most secure way of combining two keys, as compared to a single round of SHA256. hmac is designed to fold in the key material in 2 hash operations, which helps resist chosen plaintext attacks on sha-256, although SHA256 has no known chosen plaintext attacks at this time. Symmetric ciphers are considered less reliable than hashes for ...


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 ...


0

As owlstead wrote in the comments, you don't need a real public key infrastructure to use public key signatures for this. You can have A create a key-pair, then distribute the public key to B and C in advance, before any communications start. A symmetric setup isn't much more complex either. You need two secrets, one between A and B, one between A and C. ...



Top 50 recent answers are included