On 7/19/21 1:52 PM, Stephan Mueller wrote:

Am Montag, dem 19.07.2021 um 13:10 +0200 schrieb Hannes Reinecke:

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On 7/19/21 12:19 PM, Stephan Mueller wrote:

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Am Montag, dem 19.07.2021 um 11:57 +0200 schrieb Hannes Reinecke:

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On 7/19/21 10:51 AM, Stephan Mueller wrote:

[ .. ]

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Thank you for clarifying that. It sounds to me that there is no
defined protocol (or if there, I would be wondering how the code would have
worked
with a different implementation). Would it make sense to first specify
a protocol for authentication and have it discussed? I personally think
it is a bit difficult to fully understand the protocol from the code and
discuss protocol-level items based on the code.

Oh, the protocol _is_ specified:

 
https://nvmexpress.org/wp-content/uploads/NVM-Express-Base-Specification-2_0-2021.06.02-Ratified-5.pdf

It's just that I have issues translating that spec onto what the kernel
provides.

according to the naming conventions there in figures 447 and following:

- x and y: DH private key (kernel calls it secret set with dh_set_secret
or
encoded into param.key)

But that's were I got confused; one needs a private key here, but there
is no obvious candidate for it. But reading it more closely I guess the
private key is just a random number (cf the spec: g^y mod p, where y is
a random number selected by the host that shall be at least 256 bits
long). So I'll fix it up with the next round.

Here comes the crux: the kernel has an ECC private key generation function
ecdh_set_secret triggered with crypto_kpp_set_secret using a NULL key, but it
has no FFC-DH counterpart.

That said, generating a random number is the most obvious choice, but not the
right one.

The correct one would be following SP800-56A rev 3 and here either section
5.6.1.1.3 or 5.6.1.1.4.

Hmm. Okay. But after having read section 5.6.1.1.4, I still do have some
questions.

Assume we will be using a bit length of 512 for FFDHE, then we will
trivially pass Step 2 for all supported FFDHE groups (the maximum
symmetric-equivalent strength for ffdhe8192 is 192 bits).

From my understanding, the random number generator will fill out all
available bytes in the string (and nothing more), so we trivially
satisfy step 3 and 4.

And as q is always larger than the random number, step 6 reduces to
'if (c > 2^N - 2)', ie we just need to check if the random number is a
string of 0xff characters. Which hardly is a random number at all, so
it'll be impossible to get this.

Which then would mean that our 'x' is simply the random number + 1,
which arguably is slightly pointless (one more than a random number is
as random as the number itself), so I do feel justified with just
returning a random number here.

Am I wrong with that reasoning?

Cheers,

Hannes
-- 
Dr. Hannes Reinecke		           Kernel Storage Architect
hare@suse.de			                  +49 911 74053 688
SUSE Software Solutions Germany GmbH, Maxfeldstr. 5, 90409 Nürnberg
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