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Commented index_reverse and acceleration
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@ -19,6 +19,10 @@ END_PROVIDER
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subroutine two_e_integrals_index(i,j,k,l,i1)
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use map_module
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implicit none
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BEGIN_DOC
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! Gives a unique index for i,j,k,l using permtuation symmetry.
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! i <-> k, j <-> l, and (i,k) <-> (j,l)
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END_DOC
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integer, intent(in) :: i,j,k,l
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integer(key_kind), intent(out) :: i1
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integer(key_kind) :: p,q,r,s,i2
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@ -36,14 +40,25 @@ end
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subroutine two_e_integrals_index_reverse(i,j,k,l,i1)
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use map_module
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implicit none
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BEGIN_DOC
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! Computes the 4 indices $i,j,k,l$ from a unique index $i_1$.
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! For 2 indices $i,j$ and $i \le j$, we have
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! $p = i(i-1)/2 + j$.
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! The key point is that because $j < i$,
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! $i(i-1)/2 < p \le i(i+1)/2$. So $i$ can be found by solving
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! $i^2 - i - 2p=0$. One obtains $i=1 + \sqrt{1+8p}/2$
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! and $j = p - i(i-1)/2$.
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! This rule is applied 3 times. First for the symmetry of the
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! pairs (i,k) and (j,l), and then for the symmetry within each pair.
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END_DOC
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integer, intent(out) :: i(8),j(8),k(8),l(8)
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integer(key_kind), intent(in) :: i1
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integer(key_kind) :: i2,i3
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i = 0
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i2 = ceiling(0.5d0*(dsqrt(8.d0*dble(i1)+1.d0)-1.d0))
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l(1) = ceiling(0.5d0*(dsqrt(8.d0*dble(i2)+1.d0)-1.d0))
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i2 = ceiling(0.5d0*(dsqrt(dble(shiftl(i1,3)+1))-1.d0))
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l(1) = ceiling(0.5d0*(dsqrt(dble(shiftl(i2,3)+1))-1.d0))
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i3 = i1 - shiftr(i2*i2-i2,1)
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k(1) = ceiling(0.5d0*(dsqrt(8.d0*dble(i3)+1.d0)-1.d0))
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k(1) = ceiling(0.5d0*(dsqrt(dble(shiftl(i3,3)+1))-1.d0))
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j(1) = int(i2 - shiftr(l(1)*l(1)-l(1),1),4)
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i(1) = int(i3 - shiftr(k(1)*k(1)-k(1),1),4)
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