Prime Numbers

6.2 Number field sieve 287
to the left side, so (6.8) simplifies to
ln X ∼ 2
ln n.
d
Hence the running time with d fixed is
L(X) √ 2+o(1) = L(n)
√ 4/d+o(1),
which suggests that NFS will not do better than QS until we take d =5or
larger.
Now let us assume that d →∞as n →∞. Then we may replace the
coefficient d + 1 in the last term of (6.8) with d, getting
ln X ∼ 2
1
ln n + d ln X ln ln X.
d 2
Let us somewhat imprecisely change the “∼” to“=”andtrytochoosedso as to minimize X. (An optimal X0 will have the property that ln X0 ∼ ln X.)
Taking the derivative with respect to the “variable” d, wehave
X ′
X
−2
= ln n +
d2 Setting X ′ =0,weget
so that
Then
1
2 ln X ln ln X + dX′ (1 + ln ln X)
.
1 4X ln X ln ln X
d =(2lnn) 1/2 ((1/2) ln X ln ln X) −1/4 ,
ln X = 2(2 ln n) 1/2 ((1/2) ln X ln ln X) 1/4 .
(ln X) 3/4 = 2(2 ln n) 1/2 ((1/2) ln ln X) 1/4 ,
so that 3
1
4 ln ln X ∼ 2 ln ln n. Substituting, we get
or
(ln X) 3/4 ∼ 2(2 ln n) 1/2 ((1/3) ln ln n) 1/4 ,
ln X ∼ 4
3 1/3 (ln n)2/3 (ln ln n) 1/3 .
So the running time for NFS is
(64/9) 1/3 1/3 2/3
+ o(1) (ln n) (ln ln n)
.
L(X) √ 2+o(1) =exp
Thevaluesofd that achieve this heuristic asymptotic complexity satisfy
d ∼
1/3 3lnn
.
ln ln n
2