What size should the HMAC key be with SHA-256?

Question

I'm trying to generate a secret key to be used for HMAC SHA-256 signature processing. I've seen many sample of keys with variable length from 32 characters to 96 characters.

What is the ironclad rule for this key size?

Answer 1

Short answer: 32 bytes of full-entropy key is enough.

Assuming full-entropy key (that is, each bit of key is chosen independently of the others by an equivalent of fair coin toss), the security of HMAC-SHA-256 against brute force key search is defined by the key size up to 64 bytes (512 bits) of key, then abruptly drops to 32 bytes (256 bits) for larger keys; that's because in the later case, the key is hashed to 32 bytes before use. It is an argument to use a 64-byte key: it's the size giving the maximum resistance to brute force key search; and beside the key being harder to manage than a 32-byte one, using 64 bytes does not harm security, and leaves speed almost unchanged (there is no additional hashing done).

On the other hand, 256 bits of security is way more than enough for anything even vaguely foreseable, including quantum computers. If MACs are computed at a rate of $2^{88}$ per year (less hashing effort than dilapidated in bitcoin mining each year in [2018-2023]), and they could be checked among known MACs for $2^{32}$ different keys at that rate (arguably requiring more additional effort than hashing), and we wanted residual odds of $2^{-35}$ that any key is found within 32 years, 160 bits of key entropy is enough, at least ignoring CRQC.

HMAC-SHA-256 is designed for 256-bit (32-byte) cryptographic resistance in mind, with no strong argument that using a key with more entropy improves the security; beyond that, there is no assurance given by the best security proof available (Mihir Bellare: New Proofs for NMAC and HMAC: Security without Collision-Resistance, with extended abstract in Crypto 2006 Proceedings). Thus if the key is full-entropy, there is no strong argument to use a key of more then 32 bytes.

If the key is not known to be full-entropy, there is an obvious, reasonable argument that large keys are necessary. For example, if the key was a Diceware passphrase, which has an entropy of $5\log_2(6)\approx12.9$ bit/word, there needs to be 20 words in the key, that is up to 139 characters (with 6 characters per word, and space between words), to reach 256 bits of entropy.

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^{Originally an answer to this question.}

Answer 2

The only rule for the key is that it should at least contain 256 bits of randomness. If the key is smaller you may not get the full security of HMAC-SHA-256. The full security of HMAC is basically identical to the output size. Unless you are trying to protect yourself against quantum computers you should be able to get away with a key that contains 128 bits of entropy though.

Here's the text from the HMAC standard captured in RFC 2104:

The authentication key K can be of any length up to B, the block length of the hash function. Applications that use keys longer than B bytes will first hash the key using H and then use the resultant L byte string as the actual key to HMAC. In any case the minimal recommended length for K is L bytes (as the hash output length).

So preferably the entropy of the 256 bit key should be condensed into 32 bytes. What you are talking about is probably the hexadecimal representation of those 32 bytes. If the key is too large it may affect performance and efficiency of the HMAC function. Many libraries only allow binary to be inserted using octets bytes anyway, so in that case it makes sense to hex decode the key before you use it.

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HMAC uses a hash internally, which is defined for any bit string. This hash is used both for the key as for the value. So in principle you can feed it anything you want up to the maximum hash size (which you will never reach).

Definition from Wikipedia:

$ {\begin{aligned}\operatorname {HMAC} (K,m)&=\operatorname {H} {\Bigl (}{\bigl (}K'\oplus opad{\bigr )}\parallel \operatorname {H} {\bigl (}\left(K'\oplus ipad\right)\parallel m{\bigr )}{\Bigr )}\\K'&={\begin{cases}\operatorname {H} \left(K\right)&K{\text{ is larger than block size}}\\K&{\text{otherwise}}\end{cases}}\end{aligned}} $

Note the repetition of the key with regards to efficiency.

Answer 3

The concrete answer is B, as defined in RFC 2104

The authentication key K can be of any length up to B, the block length of the hash function. (B=64 for all the above mentioned examples of hash functions)

It is difficult to determine what is meant by "the block length of the hash function", so let's go digging in another RFC for information!

From RFC 4634, Page 18

SHA256_Message_Block_Size = 64

Then a lot further down, on Pages 69-70:

/*
 * USHABlockSize
 *
 * Description:
 *   This function will return the blocksize for the given SHA
 *   algorithm.
 *
 * Parameters:
 *   whichSha:
 *     which SHA algorithm to query
 *
 * Returns:
 *   block size
 *
 */
int USHABlockSize(enum SHAversion whichSha)
{
  switch (whichSha) {
    case SHA1:   return SHA1_Message_Block_Size;
    case SHA224: return SHA224_Message_Block_Size;
    case SHA256: return SHA256_Message_Block_Size;
    case SHA384: return SHA384_Message_Block_Size;
    default:
    case SHA512: return SHA512_Message_Block_Size;
  }
}

Picking anything greater than B will get hashed to L, the output of the hash function H.

B offers more security than L, so pick B.