Re: [Cfrg] Weak Diffie-Hellman Primes

[Anna Johnston <[e-mail]>]

Hi Michael,
  The attack you mentioned is logical. If I am not mistaken, the attack is recovering, bit by bit, the quotient (v) in the equation: 2^X = vP + r, where r = 2^X mod P. The fly in the attack is size. 
(1) The smallest P in this list is over 700 bits: P > 2^{700}.
(2) X is on the order of P, as 2 has order q where P=2q+1.
(3) so, 2^X, in its unreduced form, is greater than 2^{2^{700}}
(4) That means that v (the quotient) is greater than 2^{2^{700}}/2^{700}=2^{2^{700}-700} which is still about 2^{2^{700}}, or 2^{700} bits long. 

This size means that only a minuscule fraction of the quotient bits could be computed/stored (not enough seconds in time or atoms in the known universe).

Hope this helps,

Anna

> On Oct 15, 2020, at 09:15, Michael D'Errico <[e-mail]> wrote:
> 
> Hi,
> 
> I've figured out just a bit more.  I'm actually trying
> not to work on this problem, but my brain keeps
> thinking about it anyway.  The special format of
> the prime numbers in RFC's 2409, 3526, and
> 7919 allow you to create an interesting device to
> calculate the private value from a Diffie-Hellman
> public value using on the order of N bits (size of
> prime P).
> 
> The lowest 64 bits of the Diffie-Hellman public
> value Y = (2^X mod P) contain a record of when
> the mod operation performed a subtraction of a
> shifted version of P (see my original message
> quoted below).  You can reconstruct a larger value
> the modulo operation was working on by adding
> back:
> 
>    uint64 y = Y;    // low 64 bits of public value
> 
>    Y += y * P;    // mod P touched this number
> 
> This updated version of Y contains the number
> that the mod P operation was working on 64 steps
> prior to finishing.  But since the generator was 2
> (and 2^X is a solitary 1 followed by a huge string
> of zeros), this new larger value of Y will have 64
> zeros at the end which can be removed:
> 
>    Y >>= 64;    // remove 64 zero bits
> 
> This process can now repeat using the lowest 64
> bits of Y again:
> 
> repeat {
> 
>    uint64 y = Y;
> 
>    Y += y * P;
> 
>    Y >>= 64;
> 
> } until (???);
> 
> This process unwinds the modulo P operation all
> the way back to 2^X in steps of size 64.  The only
> thing missing is the terminating condition, which
> I'm pretty sure would be: Y contains a single bit
> with value 1 (somewhere in the top 64 bits).  If
> so, then the value of y would be zero a few times
> in a row and then contain a single one bit.
> 
> If you can reliably determine when the algorithm
> has finished, you could build a device which
> computes X from Y using roughly N bits.  It would
> be really slow, but it would be guaranteed to find
> the answer (32 times faster than brute force
> reversing the mod P operation one bit at a time
> on average?).
> 
> Perhaps there is a way to make this algorithm
> (much) faster?
> 
> Even if there isn't, it is very interesting to me that
> these primes exhibit such a strange feature.
> 
> Mike
> 
> 
>> From: Michael D'Errico <[e-mail]>
>> Sent: *Oct 10, 2020 6:01 PM
>> To: [e-mail]
>> Subject: [Cfrg] Weak Diffie-Hellman Primes
>> 
>> Hi,
>> 
>> I'm not a member of this list, but was encouraged to
>> start a discussion here about a discovery I made
>> w.r.t. the published Diffie-Hellman prime numbers in
>> RFC's 2409, 3526, and 7919.  These primes all have
>> a very interesting property where you get 64 or more
>> bits (the least significant bits of 2^X mod P for some
>> secret X and prime P) detailing how the modulo
>> operation was done.  These 64 bits probably reduce
>> the security of Diffie-Hellman key exchanges though
>> I have not tried to figure out how.
>> 
>> The number 2^X is going to be a single bit with value
>> 1 followed by a lot of zeros.  All of the primes in the
>> above mentioned RFC's have 64 bits of 1 in the most
>> and least significant positions.  The 2's complement
>> of these primes will have a one in the least significant
>> bit and at least 63 zeros to the left.
>> 
>> When you think about how a modulo operation is done
>> manually, you compare a shifted version of P against
>> the current value of the operand (which is initially 2^X)
>> and if it's larger than the (shifted) P, you subtract P at
>> that position and shift P to the right, or if the operand
>> is smaller than (the shifted) P, you just shift P to the
>> right without subtracting.
>> 
>> Instead of subtracting, you can add the 2's complement
>> I mentioned above.  Because of the fact that there are
>> 63 zeros followed by a 1 in the lowest position, you will
>> see a record of when the modulo operation performed
>> a subtraction (there's a one) and when it didn't (there's
>> a zero).
>> 
>> You can use the value of the result you were given by
>> your peer (which is 2^X mod P) and then add back the
>> various 2^j * P's detailed wherever the lowest 64 bits
>> had a value of 1 to find the state of the mod  P operation
>> when it wasn't yet finished.  This intermediate result is
>> likely going to make it easier to determine X than just a
>> brute force search.
>> 
>> I don't plan to join this list, though I am flattered to have
>> been asked to do so.  I'm not a cryptographer.
>> 
>> Mike
> 
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