Prime Numbers

310 Chapter 6 SUBEXPONENTIAL FACTORING ALGORITHMS
square” congruences each require O(ln β N) operations to discover. Give the
(heuristic) complexity, then, for factoring via this magic algorithm.
6.7. AGraycodeisasequenceofk-bit binary strings in such an order that
when moving from one string to the next, one and only one bit flips to its
opposite bit. Show that such a code—whether for the self-initialization QS
option or any other application—can be generated with ease, using a function
that employs exclusive-or “∧” and shift “>>” operators in the following
elegant way:
g(n) =n ∧ (n >>1).
This very simple generator is easily seen to yield, for example, a 3-bit Gray
counter that runs:
(g(0),...,g(7)) = (000, 001, 011, 010, 110, 111, 101, 100),
this counting chain clearly having exactly one bit flip on each iteration.
6.8. Show that if n ≥ 64 and m = n 1/3 ,thenn1.5(d/ ln 2) d ,andm = n 1/d ,then
n
|c| ≥ 1, where c is the leading coefficient of g(x), and similarly that
|h(m)| > 1. Thus, the factorization n = g(m)h(m) is nontrivial.
6.10. This exercise is to prove the theorem of Exercise 6.9 in the remaining
case m = 2. Hint: By a slightly more elaborate inequality as in (1) of Exercise
6.9(usingthatRe(cd−2/z) ≥ 0 for Re(z) > 0), show that every root ρ of f has
Re(ρ) < 1.49. Then let G(x) =g(x+1.49) and show that all of the coefficients