Re: [Cfrg] uniform random distribution in ECDH public key

One general advantage: after hashing the shared secret is pseudorandom.  It has less entropy but it is harder to distinguish from a uniformly random bit string.   This may matter if one just wanted to use the key to pad a message.  This does not really matter for AES, but the ECDH standards don't specify how the key is, so have to be general.

A second advantage, nonces and identifiers can be thrown into the hash input, for various gains, such as generating several keys, one for AES and for HMAC.

A third advantage, if the shared key is later revealed, then the hash prevents the calculation of the raw ECDH value, which is harmful mostly in the case where one public key is static: it allows the adversary to do replay, invalid curve attacks, and Cheon-type attacks.

----- Original Message -----
From: David Jacobson [mailto:dmjacobson@sbcglobal.net]
Sent: Thursday, August 23, 2012 01:17 AM
To: cfrg@irtf.org <cfrg@irtf.org>
Subject: Re: [Cfrg] uniform random distribution in ECDH public key

On 08/14/2012 11:23 AM, Dan Harkins wrote:
>    Hi Bob,
>
> On Tue, August 14, 2012 11:01 am, Robert Moskowitz wrote:
>> I understand from RFC 6090 and 5869 that the secret key produced from an
>> ECDH exchange is not uniformly randomly distributed and that is why we
>> have the 'Extract' phase in HKDF.  Got that.
>>
>> This question is about the public key, g^j:
>>
>> I understand that like j, it must be a point on the curve, thus if the
>> curve is p-256, both j and g^j are 256 bits long.  But is g^j uniformly
>> randomly distributed like j is suppose to be?
>    No, it's not. It's it's a special pair (x,y) that satisfy the equation
> of the
> curve:  y^2 = x^3 + ax + b. Not all pairs will satisfy that equation. I
> believe about half of them will and about half won't.
>
>    For x to be random, each number between 0 and p would have equal
> probability. But that's not the case since about half won't.
>
>> Side question:  I am still unclear on the length of the exchanged secret
>> (g^j)^k, is it 256 bits (for p-256) or larger (perhaps 512 bits)?
>    The result of an ECDH is an element in the group so it's also an (x,y)
> pair but the secret that you use in your KDF is the x coordinate of that
> result. The y coordinate is discarded.
>
>    regards,
>
>    Dan.
>
>
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So now that we are into tutorial mode on this, I'd like to ask a 
question.  Standard procedure for Diffie-Hellman key exchange is to 
construct the session key from the X-coordinate by hashing.  Now suppose 
that I'm using the NIST P-256 curve and the symmetric encryption 
functions is AES-256.

The number of possible shared key values is the order of the curve - 1 
(point at infinity isn't used), which is extremely close to 2^256.    
These points come in pairs, if there is a point at an X value, there are 
2, one at Y and the other -Y.  So essentially all X values occur with 
probability  very close to 2/2^256, which means that the X-coordinate 
after the DH procedure can be thought of as a source with 255 bits of  
min-entropy.  If we hash the X coordinate with SHA-256, we actually lose 
a little bit of entropy, since some X values will collide and produce 
some session key with probability higher than 2/2^256, lowering the 
min-entropy.

So what is the advantage of the hash operation?

Thank you,

     --David Jacobson

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