I need cryptography advice regarding this issue.

Kamus is a service that encrypts secrets for applications running on Kubernetes. When using AES (actually, Rijndael) symmetric encryption, Kamus uses a single key to encrypt the secrets for all apps.

Let's say I want to create from the generic key, a key that is unique per application - so that secrets that are encrypted for a specific application will not be able to decrypt with the key of another application (this is the case when using a KMS). I'm looking for an algorithm to "derive" a key from the application name (the service account name, to be accurate), something like:

f(key, name) = key

Is there an algorithm that I can use? What is the best approach here?

  • 3
    $\begingroup$ HKDF(key + name, digest_len) $\endgroup$ – SamG101 Dec 22 '19 at 19:50
  • $\begingroup$ @SamG101 HKDF not defined as you listed. If we combine extract and expand then HKDF(IKM , Salt, info, length) $\endgroup$ – kelalaka Dec 23 '19 at 9:24
  • 3
    $\begingroup$ Ohh right yes you're right, I was just suggesting the concept $\endgroup$ – SamG101 Dec 23 '19 at 9:26

HMAC-based Key Derivation Function (HKDF) rfc5869 is what you are looking for. HMAC security proof uses the fact that the compression function of the underlying hash is itself a PRF.

HKDF follows the "extract-then-expand" paradigm, where the KDF logically consists of two modules. The first stage takes the input keying material and "extracts" from it a fixed-length pseudorandom key K. The second stage "expands" the key K into several additional pseudorandom keys (the output of the KDF).


$$\text{HKDF-Extract}(salt, IKM) \to PRK,$$where PRK is a pseudorandom key.

If the Input Key Material (IKM) is already a random key, as in your case, then the extract is not necessary, Expand is enough. HKDF can be used without the salt, however, using salt adds to the strengthening of HKDF and supporting source-independent extraction. Two different salts with same $IKM$ result in fundamentally two different PRKs. And, in general, $x$ different salts with same $IKM$ result in fundamentally $x$ different PRKs.


$$\text{KDF-Expand}(PRK, info, L) \to OKM,$$where OKM is Output Keying Material. L is the desired key length.

The info can be used for the application-specific tag to derive different keys.

$$\text{KDF-Expand}(\text{Inittal Key}, \text{"application 1"}, 128) \to OKM_1$$ $$\text{KDF-Expand}(\text{Inittal Key}, \text{"application 2"}, 256) \to OKM_2$$

Note: There is also old KDF1 and KDF2

$$K_i = \operatorname{KDF}(K_{master}, i) = \operatorname{H}(K_{master} \mathbin\| c)$$ where $c$ is 4 byte encoded $i$, and it was commonly used wiht MD5,SHA-1, and SHA-256.

A sample code with python hkdf that contains the pop count on the output's x-or;

from hkdf import hkdf_extract,hkdf_expand
from binascii import unhexlify,b2a_hex
import sys

def bxor(b1, b2): # use xor for bytes
    parts = []
    for b1, b2 in zip(b1, b2):
        parts.append(bytes([b1 ^ b2]))
    return b''.join(parts)

prk = hkdf_extract(unhexlify(b"8e94ef805b93e683ff18"), b"asecretpassword")
key1 = hkdf_expand(prk, b"application 1", 16)
key2 = hkdf_expand(prk, b"application 2", 16)

print (b2a_hex(key1))
print (b2a_hex(key2))

#count the number of differnt bits by x-or and popup count.
print (bin(int.from_bytes(bxor(key1,key2), byteorder=sys.byteorder))[2:].count('1'))


  • 3
    $\begingroup$ Thanks! This is exactly what I'm looking for :) $\endgroup$ – Omer Levi Hevroni Dec 23 '19 at 6:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.