Dear Julian,
Good question! The techniques from the Nguyen-Shparlinski cocyclic
lattices paper should extend to this case, but we don't have the
result yet. We will look into it.
We can however carry out the counting argument in the special case of a
squarefree determinant D (which is the case that occurs in Squirrels).
Note that lattices with such a squarefree determinant are all cocyclic
since the only abelian group of order D is Z/DZ.
To count the lattices of determinant D, it suffices to count the HNFs. In
such an HNF, each of the prime factors p of D will occur at exactly one
position i_p on the diagonal, and the number of HNFs with the pattern
(i_p)_{p|D} is \prod_{p|D} p^{i_p-1}. Therefore, the total number of
lattices of determinant D is exactly:
N(D) = \prod_{p|D} (1 + p + ··· + p^{n-1}) = \prod_{p|D} (p^n-1)/(p-1)
and of those, exactly D^{n-1} have an HNF of the form we desire (namely
with diagonal (1,...,1,D)). The proportion is therefore:
D^{n-1} / N(D) = \prod_{p|D} p^{n-1}(p-1) / (p^n-1)
which is always greater than phi(D)/D and converges to that value as
n->\infty.
In our case, we fix D as a product of distinct primes that are moderately
large (31 bits), which makes the probability of having the required HNF
shape very close to 1. For example, it is above 1-2^{-23} for parameter
set Squirrels-I (in other words, a random lattice of determinant D will
be rejected in key generation with probability < 2^{-23}). The situation
would be different with D divisible by many small primes. However, the
arguments of Paz and Schnorr saying that all lattices are arbitrarily
close to cocylic lattices still hold when we consider only cocyclic
lattices whose determinant does not have small prime factors.
Anyway, we will try to carry out the density computation in the style of
Nguyen-Shparlinski and send an update accordingly.
Best regards,
--
The Squirrels team.
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