Otherwise,
with
s=v.size()
and
n=end−begin,
fills the supplied range
[begin,end)
according to the following algorithm
in which
each operation is to be carried out modulo
232,
each indexing operator applied to
begin is to be taken modulo
n,
and
T(x) is defined as
x xor(x rshift27):
By way of initialization,
set each element of the range to the value
0x8b8b8b8b. Additionally,
for use in subsequent steps,
let p=(n−t)/2
and let q=p+t,
where
t=(n≥623) ? 11 : (n≥68) ? 7 : (n≥39) ? 5 : (n≥7) ? 3 : (n−1)/2;
With
m as the larger of
s+1 and
n,
transform the elements of the range:
iteratively for
k=0,…,m−1,
calculate values
r1=1664525⋅T(begin[k]xorbegin[k+p]xorbegin[k−1])r2=r1+⎧⎪⎨⎪⎩s, k=0kmodn+v[k−1], 0<k≤skmodn, s<k
and, in order,
increment
begin[k+p] by
r1,
increment
begin[k+q] by
r2,
and
set
begin[k] to
r2.Transform the elements of the range again,
beginning where the previous step ended:
iteratively for
k=m,…,m+n−1,
calculate values
r3=1566083941⋅T(begin[k]+begin[k+p]+begin[k−1])r4=r3−(kmodn)
and, in order,
update
begin[k+p] by xoring it with
r3,
update
begin[k+q] by xoring it with
r4,
and
set
begin[k] to
r4.