Prime Numbers

312 Chapter 6 SUBEXPONENTIAL FACTORING ALGORITHMS
testing your QS implementation in earnest, it will demonstrate how 12digit
factors can be extracted with QS in a matter of seconds or minutes
(depending on the efficiency of the sieve and the matrix package). This is
still somewhat slower than sheer sieving or say Pollard-rho methods, but
of course, QS can be pressed much further, with its favorable asymptotic
behavior.
(4) Try factoring the repunit
n = 1029 − 1
9
= 11111111111111111111111111111
using a forced parameter B = 40000, for which matrices will be about
2000 × 2000 in size.
(5) If you have not already for the above, implement Algorithm 6.1.1 in fast,
compiled fashion to attempt factorization of, say, 100-digit composites.
6.15. In the spirit of Exercise 6.14, we here work through the following
explicit examples of the NFS Algorithm 6.2.5. Again the point is to give the
reader some guidance and means for algorithm debugging. We shall find that
a particular obstruction—the square-rooting in the number field—begs to be
handled in different ways, depending on the scale of the problem.
(1) Start with the simple choice n = 10403 and discover that the polynomial
f is reducible, hence the very Step [Setup] yields a factorization, with no
sieving required.
(2) Use Algorithm 6.2.5 with initialization parameters as is in the pseudocode
listing, to factor n = F5 = 232 + 1. (Of course, the SNFS likes this
composite, but the exercise here is to get the general NFS working!) From
the initialization we thus have d =2,B = 265, m = 65536, k = 96,
and thus matrix dimension V = 204. The matrix manipulations then
accrue exactly as in Exercise 6.14, and you will obtain a suitable set
S of (a, b) pairs. Now, for the small composite n in question (and the
correspondingly small parameters) you can, in Step [Square roots], just
multiply out the product
(a,b)∈S
(a − bα) to generate a Gaussian integer,
because the assignment α = i is acceptable. Note how one is lucky for
such (d = 2) examples, in that square-rooting in the number field is a
numerical triviality. In fact, the square root of a Gaussian integer c + di
can be obtained by solving simple simultaneous relations. So for such small
degree as d = 2, the penultimate Step [Square roots] of Algorithm 6.2.5 is
about as simple as can be.
(3) As a kind of “second gear” with respect mainly to the square-root obstacle,
try next the same composite n = F5 but force parameters d =4,B = 600,
which choices will result in successful NFS. Now, at the Step [Square
roots], you can again just multiply out the product of terms (a − bα)
where now α = √ i, and you can then take the square root of the resulting
element
s0 + s1α + s2α 2 + s3α 3