On 10/12/19 7:23 pm, Ard Biesheuvel wrote:
> On Tue, 10 Dec 2019 at 12:04, Keerthy <j-keerthy@xxxxxx> wrote:
>>
>>
>>
>> On 10/12/19 3:37 pm, Ard Biesheuvel wrote:
>>> On Tue, 10 Dec 2019 at 11:06, Keerthy <j-keerthy@xxxxxx> wrote:
>>>>
>>>>
>>>>
>>>> On 10/12/19 3:31 pm, Ard Biesheuvel wrote:
>>>>> Hello Keerthy,
>>>>>
>>>>> On Tue, 10 Dec 2019 at 10:35, Keerthy <j-keerthy@xxxxxx> wrote:
>>>>>>
>>>>>> Hi Ard,
>>>>>>
>>>>>> I am not sure if am the first one to report this. It seems like
>>>>>> aes_expandkey is giving me different expansion over what i get with the
>>>>>> older crypto_aes_expand_key which was removed with the below commit:
>>>>>>
>>>>>> commit 5bb12d7825adf0e80b849a273834f3131a6cc4e1
>>>>>> Author: Ard Biesheuvel <ard.biesheuvel@xxxxxxxxxx>
>>>>>> Date: Tue Jul 2 21:41:33 2019 +0200
>>>>>>
>>>>>> crypto: aes-generic - drop key expansion routine in favor of library
>>>>>> version
>>>>>>
>>>>>> The key that is being expanded is from the crypto aes(cbc) testsuite:
>>>>>>
>>>>>> }, { /* From NIST SP800-38A */
>>>>>> .key = "\x8e\x73\xb0\xf7\xda\x0e\x64\x52"
>>>>>> "\xc8\x10\xf3\x2b\x80\x90\x79\xe5"
>>>>>> "\x62\xf8\xea\xd2\x52\x2c\x6b\x7b",
>>>>>> .klen = 24,
>>>>>>
>>>>>>
>>>>>> The older version crypto_aes_expand_key output that passes the cbc(aes)
>>>>>> decryption test:
>>> ...
>>>>>>
>>>>>> The difference is between 52nd index through 59.
>>>>>>
>>>>>> Any ideas if this is expected?
>>>>>>
>>>>>
>>>>> Yes, this is expected. This particular test vector uses a 192 bit key,
>>>>> so those values are DC/ignored.
>>>>
>>>> Thanks for the quick response. However with the new implementation
>>>> decryption test case fails for me with wrong result.
>>>
>>> Can you share more details please? Platform, endianness, etc ..
>>
>> Ard,
>>
>> I am trying to get aes working on a yet to be upstream TI HW crypto
>> Accelerator SA2UL. It is little endian.
>>
>> I had posted a series earlier this year:
>>
>> https://lkml.org/lkml/2019/6/28/20
>>
>> The device expects the inverse key for decryption.
>>
>
> Could you elaborate? There is no such thing as an inverse *key*, only
> an inverse *key schedule* which is used for the Equivalent Inverse
> Cipher.
Sorry i was a bit unclear.
>
> AES-192 expands the 24 byte key to 13 round keys consisting of 4
> 32-bit words each, and so the algorithm does not actually use the
> contents of slots 52 and up in this case.
>
>> In the earlier working version i was copying the ctx.key_enc[48] to
>> ctx.key_enc[53] index of the ctx.key_enc array as the 24 bytes of
>> decryption key to my hardware.
>>
>> Now as told earlier the 52nd & 53rd words are changed and hence i end up
>> in wrong result.
>>
>> Fail:
>>
>> ctx.key_dec[48] = 0xf7b0738e & ctx.key_enc[48] = 0x6fa08be9
>> ctx.key_dec[49] = 0x52640eda & ctx.key_enc[49] = 0x3c778c44
>> ctx.key_dec[50] = 0x2bf310c8 & ctx.key_enc[50] = 0x472cc8e
>> ctx.key_dec[51] = 0xe5799080 & ctx.key_enc[51] = 0x2220001
>> ctx.key_dec[52] = 0x13eaf950 & ctx.key_enc[52] = 0x13eaf850
>> ctx.key_dec[53] = 0xffff8000 & ctx.key_enc[53] = 0xffff8000
>>
>> Pass:
>>
>> ctx.key_dec[48] = 0xf7b0738e & ctx.key_enc[48] = 0x6fa08be9
>> ctx.key_dec[49] = 0x52640eda & ctx.key_enc[49] = 0x3c778c44
>> ctx.key_dec[50] = 0x2bf310c8 & ctx.key_enc[50] = 0x472cc8e
>> ctx.key_dec[51] = 0xe5799080 & ctx.key_enc[51] = 0x2220001
>> ctx.key_dec[52] = 0x105127e8 & ctx.key_enc[52] = 0x68342d29
>> ctx.key_dec[53] = 0xffff8000 & ctx.key_enc[53] = 0xddd31195
>>
>
> The old code does the following for AES-192
>
> #define loop6(i) do { \
> t = ror32(t, 8); \
> t = ls_box(t) ^ rco_tab[i]; \
> t ^= ctx->key_enc[6 * i]; \
> ctx->key_enc[6 * i + 6] = t; \
> t ^= ctx->key_enc[6 * i + 1]; \
> ctx->key_enc[6 * i + 7] = t; \
> t ^= ctx->key_enc[6 * i + 2]; \
> ctx->key_enc[6 * i + 8] = t; \
> t ^= ctx->key_enc[6 * i + 3]; \
> ctx->key_enc[6 * i + 9] = t; \
> t ^= ctx->key_enc[6 * i + 4]; \
> ctx->key_enc[6 * i + 10] = t; \
> t ^= ctx->key_enc[6 * i + 5]; \
> ctx->key_enc[6 * i + 11] = t; \
> } while (0)
>
> case AES_KEYSIZE_192:
> ctx->key_enc[4] = get_unaligned_le32(in_key + 16);
> t = ctx->key_enc[5] = get_unaligned_le32(in_key + 20);
> for (i = 0; i < 8; ++i)
> loop6(i);
> break;
>
> so while it happens to populate slots 52 and 53 as well (when i == 7),
> the AES spec does not actually cover this, given that those values are
> not actually used in the computation (and I am at a loss understanding
> why it should make a difference in your case).
>
> In any case, you can work around this by calculating the missing
> values in your driver's expand_key() routine,
>
> ctx.key_enc[52] = ctx.key_enc[51] ^ ctx.key_enc[46];
> ctx.key_enc[53] = ctx.key_enc[52] ^ ctx.key_enc[47];
Thanks a lot for the detailed explanation.
Thanks a lot Ard. The above work around gets it working on my hardware
for 24 byte keys. I will dig more details on that.
Best Regards,
Keerthy
>
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