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Re: How do I choose constants suitable for Diffe-Hellman?
[email protected] writes:
>How do I choose constants suitable for Diffe-Hellman?
>According to _Applied Cryptography_ n should be prime,
>also (n-1)/2 should also be prime. g should be a primitive
>root of unity mod n. n should be 512 or 1024 bits long.
>Are there any other requirements?
These requirements are slightly overkill, IMO. n does have to be prime,
but what you really want is to have g generate a "large enough" sub-group
of the numbers from 1 to n. One way to achive this is to have (n-1)/2
also be prime, in which case the order of g (the length of g^0,g^1,...,1)
is either 1, n-1, 2, or (n-1)/2. The odds of it being 1 or 2 are
practically nil, so you could really use a random g since a period of
(n-1)/2 is more than good enough. Or, you could test g by raising it to
the (n-1)/2 power and if the answer is 1 reject it and try another g.
That way you get one with period n-1 which is maximal.
There was a program posted here last time we discussed this (maybe four
months ago?) which sieved for both n prime and (n-1)/2 prime. It was
One thing you can do which IMO is just as good is to choose a g with a
considerably smaller period. There are two known ways to solve
discrete logs; one depends on the size of n and the other depends on
the size of the order of g(|g|). The second one is much weaker so if
you choose the size of |g| to provide about as much security as the
method based on the size of n you get something like n=512, |g|=140.
This is used in the DSS, I believe.
The advantage of this is that it is faster to exponentiate g^x in DH
since x will be only 140 bits.
So, to use this, pick a prime q of 140 bits, then find a prime n equal to
kq+1 for some k, such that n is 512 bits. This assures that there are
some generators g which have a period of q. There is an easy trick to
find one: pick a random number a < n, and set g = a ^ ((n-1)/q). It
follows that g^q equals 1 (since it is a^(n-1)), and since q is prime it
must be the order of g.
As I said, you can always use the full DH, but you would be in good
company using the small-q version. One question is the size of q to use
for n=1024. I haven't seen a clear answer to that, but the general
principle is that if solving discrete logs becomes X times harder, you
should increase q by a factor of X^2. So if DH is a million times harder
for n=1024 than for n=512 (it's hard to tell with all of the O(1) factors
in the formulas) then q should be 40 bits longer or about 180 bits.