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Commented index_reverse and acceleration

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Anthony Scemama 2019-04-10 17:19:44 +02:00
parent 8deb6ae0d7
commit 178a8f5760
1 changed files with 18 additions and 3 deletions
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21
src/ao_two_e_ints/map_integrals.irp.f
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@ -19,6 +19,10 @@ END_PROVIDER
subroutine two_e_integrals_index(i,j,k,l,i1)
use map_module
implicit none
BEGIN_DOC
! Gives a unique index for i,j,k,l using permtuation symmetry.
! i <-> k, j <-> l, and (i,k) <-> (j,l)
END_DOC
integer, intent(in) :: i,j,k,l
integer(key_kind), intent(out) :: i1
integer(key_kind) :: p,q,r,s,i2
@ -36,14 +40,25 @@ end
subroutine two_e_integrals_index_reverse(i,j,k,l,i1)
use map_module
implicit none
BEGIN_DOC
! Computes the 4 indices $i,j,k,l$ from a unique index $i_1$.
! For 2 indices $i,j$ and $i \le j$, we have
! $p = i(i-1)/2 + j$.
! The key point is that because $j < i$,
! $i(i-1)/2 < p \le i(i+1)/2$. So $i$ can be found by solving
! $i^2 - i - 2p=0$. One obtains $i=1 + \sqrt{1+8p}/2$
! and $j = p - i(i-1)/2$.
! This rule is applied 3 times. First for the symmetry of the
! pairs (i,k) and (j,l), and then for the symmetry within each pair.
END_DOC
integer, intent(out) :: i(8),j(8),k(8),l(8)
integer(key_kind), intent(in) :: i1
integer(key_kind) :: i2,i3
i = 0
i2 = ceiling(0.5d0*(dsqrt(8.d0*dble(i1)+1.d0)-1.d0))
l(1) = ceiling(0.5d0*(dsqrt(8.d0*dble(i2)+1.d0)-1.d0))
i2 = ceiling(0.5d0*(dsqrt(dble(shiftl(i1,3)+1))-1.d0))
l(1) = ceiling(0.5d0*(dsqrt(dble(shiftl(i2,3)+1))-1.d0))
i3 = i1 - shiftr(i2*i2-i2,1)
k(1) = ceiling(0.5d0*(dsqrt(8.d0*dble(i3)+1.d0)-1.d0))
k(1) = ceiling(0.5d0*(dsqrt(dble(shiftl(i3,3)+1))-1.d0))
j(1) = int(i2 - shiftr(l(1)*l(1)-l(1),1),4)
i(1) = int(i3 - shiftr(k(1)*k(1)-k(1),1),4)