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first complex reverse compound index function

This commit is contained in:
Kevin Gasperich 2020-01-30 16:53:49 -06:00
parent aac2c60971
commit 5f37d50f23

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@ -159,6 +159,75 @@ subroutine two_e_integrals_index_reverse(i,j,k,l,i1)
end end
subroutine two_e_integrals_index_reverse_periodic_1(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.
! always returns first set such that i<=k, j<=l, ik<=jl
END_DOC
integer, intent(out) :: i(4),j(4),k(4),l(4)
integer(key_kind), intent(in) :: i1
integer(key_kind) :: i2,i3
i = 0
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(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)
!ijkl a+ib
i(2) = j(1) !jilk a+ib
j(2) = i(1)
k(2) = l(1)
l(2) = k(1)
i(3) = i(1) !ilkj a-ib
j(3) = l(1)
k(3) = k(1)
l(3) = j(1)
i(4) = l(1) !lijk a-ib
j(4) = i(1)
k(4) = j(1)
l(4) = k(1)
integer :: ii, jj
do ii=2,4
do jj=1,ii-1
if ( (i(ii) == i(jj)).and. &
(j(ii) == j(jj)).and. &
(k(ii) == k(jj)).and. &
(l(ii) == l(jj)) ) then
i(ii) = 0
exit
endif
enddo
enddo
! This has been tested with up to 1000 AOs, and all the reverse indices are
! correct ! We can remove the test
! do ii=1,8
! if (i(ii) /= 0) then
! call two_e_integrals_index(i(ii),j(ii),k(ii),l(ii),i2)
! if (i1 /= i2) then
! print *, i1, i2
! print *, i(ii), j(ii), k(ii), l(ii)
! stop 'two_e_integrals_index_reverse failed'
! endif
! endif
! enddo
end