Fastest way to get Inverse Mapping from Values to Indices in fortran - arrays

Question

Here's a array A with length N, and its values are between 1 and N (no duplication).
I want to get the array B which satisfies B[A[i]]=i , for i in [1,N]
e.g.
for A=[4,2,1,3], I want to get
B=[3,2,4,1]
I've writen a fortran code with openmp as showed below, array A is given by other procedure. For N = 1024^3(~10^9), it takes about 40 seconds, and assigning more threads do little help (it takes similar time for OMP_NUM_THREADS=1, 4 or 16). It seens openmp does not work well for very large N. (However it works well for N=10^7)
I wonder if there are other better algorithm to do assignment to B or make openmp valid.
the code:
subroutine fill_inverse_array(leng, A, B)
use omp_lib
implicit none
integer*4 intent(in) :: leng, i
integer*4 intent(in) :: A(leng)
integer*4 intent(out) :: B(leng)
!$omp parallel do private(i) firstprivate(leng) shared(A, B)
do i=1,leng
B(A(i))=i
enddo
!$omp end parallel do
end subroutine

Answer 1

It's a slow day here so I ran some tests. I managed to squeeze out a useful increase in speed by rewriting the expression inside the loop, from B(A(i))=i to the equivalent B(i) = A(A(i)). I think this has a positive impact on performance because it is a little more cache-friendly.
I used the following code to test various alternatives:
A = random_permutation(length)
CALL system_clock(st1)
B = A(A)
CALL system_clock(nd1)
CALL system_clock(st2)
DO i = 1, length
B(i) = A(A(i))
END DO
CALL system_clock(nd2)
CALL system_clock(st3)
!$omp parallel do shared(A,B,length) private(i)
DO i = 1, length
B(i) = A(A(i))
END DO
!$omp end parallel do
CALL system_clock(nd3)
CALL system_clock(st4)
DO i = 1, length
B(A(i)) = i
END DO
CALL system_clock(nd4)
CALL system_clock(st5)
!$omp parallel do shared(A,B,length) private(i)
DO i = 1, length
B(A(i)) = i
END DO
!$omp end parallel do
CALL system_clock(nd5)
As you can see, there are 5 timed sections in this code. The first is a simple one-line revision of your original code, to provide a baseline. This is followed by an unparallelised and then a parallelised version of your loop, rewritten along the lines I outlined above. Sections 4 and 5 reproduce your original order of operations, first unparallelised, then parallelised.
Over a series of four runs I got the following average times. In all cases I was using arrays of 10**9 elements and 8 threads. I tinkered a little and found that using 16 (hyperthreads) gave very little improvement, but that 8 was a definite improvement on fewer. Some average timings
Sec 1: 34.5s
Sec 2: 32.1s
Sec 3: 6.4s
Sec 4: 31.5s
Sec 5: 8.6s
Make of those numbers what you will. As noted above, I suspect that my version is marginally faster than your version because it makes better use of cache.
I'm using Intel Fortran 14.0.1.139 on a 64-bit Windows 7 machine with 10GB RAM. I used the '/O2' option for compiler optimisation.