5
$\begingroup$

I want to get the original encryption key (32 byte) of AES 256 (CBC) implementation. The implementation has the values of the inversed round keys values hard coded into the decryption function. I have the 15 rounds*32 byte decryption keys Is there an algorithm that I can use to get the original generating key? I know that I can reverse the expansion of the encryption key, but the round keys for decryption has the addition "inversion" generated by this:

// inverse-cipher-ify the decryption round key (fips-197 section 5.3)
    for (var r = 1; r < rounds; r++) {
        for (var c = 0; c < 4; c++) {
            tt = this._Kd[r][c];
            this._Kd[r][c] = (U1[(tt >> 24) & 0xFF] ^
                              U2[(tt >> 16) & 0xFF] ^
                              U3[(tt >>  8) & 0xFF] ^
                              U4[ tt        & 0xFF]);
        }
    }

Can this be reverted too? Thank you

$\endgroup$
1
  • 1
    $\begingroup$ Note that the last round key used in decryption should be, in fact, the original AES key. To address your question it's not clear in your question what kind of table "this._Kd" is. Could you link to the implementation ? $\endgroup$
    – Ruggero
    Commented May 24, 2019 at 11:46

1 Answer 1

3
+50
$\begingroup$

Yes, in fact it is quite simple. You are looking at the round subkeys for what is called the equivalent inverse cipher. The decryption round subkeys are generated by performing the InverseMixColumns operation on the encryption round key, the exact same operation used by the decryption function. The code you posted uses a series of byte table lookups (U1 to U4) and XORs to perform the InverseMixColumns operation. Note that the round subkeys are 128-bits or 16 bytes, as that is the size of the block.

In order to convert the decryption round subkey into the encryption one, all you need to do is perform the forward MixColumns operation on the subkeys, except for the first and last round keys, as these are whitening subkeys and are not changed. And of course you need to use them all in reverse order.

$\endgroup$
1
  • $\begingroup$ Also, after you have derived the encryption round keys from the decryption round keys, the original N-bit key is just the first N/32 words of that. In the case of AES-128, it's the last decryption round key and you don't need MixColumns at all. $\endgroup$
    – benrg
    Commented Aug 26, 2019 at 21:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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