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# do i = 1, np {
# do j = 1, i {
# ij = ia(i) + j
# a = 1. / (i + j)
# do k = 1, i {
# maxl = k
# if (k == i) then
# maxl = j
# endif
# do l = 1, maxl {
# kl = ia(k) + l
# b = 1. / (k + l)
# val = a + b
# if (i == j) then
# val = val * .5
# endif
# if (k == l) then
# val = val * .5
# endif
# x(ij,kl) = val
# x(kl,ij) = val
# }
# }
# }
# }
#!
#
#
# As far as I remember it was the dependence test between X(IJ,KL) and X(IJ,KL).
#
# Pips discovered the following precondition:
#
{[d1,d2,d3,d4] : exists ( NP,
I,J,K,L,IJ,KJ,MAXL,
I',J',K',L',IJ',KJ',MAXL' :
I' = I+d1 &&
J' = J+d2 &&
K' = K+d3 &&
L' = L+d4
&& 1<=L && 1<=J && MAXL<=K && I<=NP && NP<=40 && 10<=NP
&& 10+8K+MAXL<=NP+8I+J && 38K+MAXL<=38I+J
&& J+K<=I+MAXL && L<=MAXL
&& 1 <= J,K <= I <= NP && 1 <=L <= MAXL
&& 1<=L' && 1<=J' && MAXL'<=K' && I'<=NP && NP<=40 && 10<=NP
&& 10+8K'+MAXL'<=NP+8I'+J' && 38K'+MAXL'<=38I'+J'
&& J'+K'<=I'+MAXL' && L'<=MAXL'
&& 1 <= J',K' <= I' <= NP && 1 <=L' <= MAXL'
)};
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