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subroutine permuto(n,n3,n4,c,f,d, crit,loc3n,work4n)
implicit doubleprecision (a-h,o-z)
dimension c(n,n),f(n,n),d(n,n),loc3n(n3),work4n(n4)
dimension bp(7),bpn(7),bpb(7)
c common /hgwc/ bp,bpn,bpb,rfnk,rfnk1,rfnk2,k,n1,n2,n3
crit=0.d0
do 1,i=1,n3
loc3n(i)=0
1 continue
nc=n
nf=n
nd=n
call hgw(n, c, f, d, nc, nf, nd, loc3n, work4n,
+bp,bpn,bpb,rfnk,rfnk1,rfnk2,k,n1,n2,n3)
crit=work4n(1)
return
end
subroutine hgw(n,c,f,d,nc,nf,nd,loc3n,work4n,
+bp,bpn,bpb,rfnk,rfnk1,rfnk2,k,n1,n2,n3)
implicit doubleprecision (a-h,o-z)
dimension c(nc,1),f(nf,1),d(nd,1),loc3n(1),work4n(1)
dimension bp(7),bpn(7),bpb(7)
ibest=1
jbest=1
c
if (n.le.1) return
n1 = n
n2 = n + n
n3 = 3*n
c initialize recurrence data
call initrd(c,f,d,nc,nf,nd,pnorm,loc3n(n1+1),loc3n(n2+1),
* work4n(1), work4n(n1+1), work4n(n2+1), work4n(n3+1),
+bp,bpn,bpb,rfnk,rfnk1,rfnk2,k,n1,n2,n3)
obj = (bp(3)*rfnk1+bp(5))*rfnk
nm1 = n - 1
c
do 30 kp1=1,nm1
k = kp1 - 1
c select candidates for position kp1 in s' and l' to give a
c maximal reduction in the average objective function.
delbst = pnorm
c delbst need only be larger than the rounding error in delta
do 20 i=kp1,n
is = n1 + i
ibar = loc3n(is)
c
do 10 j=kp1,n
jl = n2 + j
c pre-evaluate effect on obj of assignment ibar<->loc3n(jl)
temp = delta(ibar,loc3n(jl),c,f,d,nc,nf,nd,loc3n(n1+1),
* loc3n(n2+1),work4n(1),work4n(n1+1),work4n(n2+1),
* work4n(n3+1),
+bp,bpn,bpb,rfnk,rfnk1,rfnk2,k,n1,n2,n3)
if (temp.ge.delbst) go to 10
c got an improvement,so save it
ibest = i
jbest = j
delbst = temp
bpb(1) = bpn(1)
bpb(2) = bpn(2)
bpb(3) = bpn(3)
bpb(4) = bpn(4)
bpb(5) = bpn(5)
bpb(6) = bpn(6)
bpb(7) = bpn(7)
10 continue
c
20 continue
c exchange best items into position kp1
kp1s = n1 + kp1
kp1l = n2 + kp1
ibests = n1 + ibest
jbestl = n2 + jbest
itemp = loc3n(kp1s)
loc3n(kp1s) = loc3n(ibests)
loc3n(ibests) = itemp
jtemp = loc3n(kp1l)
loc3n(kp1l) = loc3n(jbestl)
loc3n(jbestl) = jtemp
obj = obj + delbst
c update recurrence data
if (kp1.ne.nm1) call updrd(f, d, nf, nd, loc3n(n1+1),
* loc3n(n2+1), work4n(1), work4n(n1+1), work4n(n2+1),
* work4n(n3+1),
+bp,bpn,bpb,rfnk,rfnk1,rfnk2,k,n1,n2,n3)
30 continue
c
c the final iteration, with k=n-1, is not performed since there is only one
c candidate pair and it is already in place. now unscramble the permutation.
c
do 40 i=1,n
il = n2 + i
is = n1 + i
ls = loc3n(is)
loc3n(ls) = loc3n(il)
40 continue
c the construction phase is now complete. the remaining code tries to
c improve the current approximate solution using pair exchanges.if the
c user wishes to trade solution quality for speed of computation, the
c following statements up to and including label 90 may be omitted.
do 70 m=1,n
c at most n times
delbst = 0.d0
do 60 i=2,n
im1 = i - 1
c
do 50 j=1,im1
temp = deltx(i,j,c,f,d,nc,nf,nd,loc3n,
+bp,bpn,bpb,rfnk,rfnk1,rfnk2,k,n1,n2,n3)
if (temp.ge.delbst) go to 50
c got an improvement,so save it
ibest = i
jbest = j
delbst = temp
50 continue
60 continue
if (delbst.ge.0.d0) go to 80
c perform exchange
li = loc3n(ibest)
lj = loc3n(jbest)
loc3n(ibest) = lj
loc3n(jbest) = li
obj = obj + delbst
70 continue
c
80 continue
work4n(1) = obj
return
end
doubleprecision function delta(ibar,jbar,c,f,d,nc,nf,nd,ls,ll,w1,
+ w2,w3,w4,bp,bpn,bpb,rfnk,rfnk1,rfnk2,k,n1,n2,n3)
implicit doubleprecision (a-h,o-z)
dimension c(nc,1),f(nf,1),d(nd,1),ls(1),ll(1),w1(1),w2(1),
* w3(1), w4(1)
dimension bp(7),bpn(7),bpb(7)
c called by the quadratic assignment heuristic hgw. returns the
c change in obj (new - old) that would be produced by transferring
c ibar from s to s' and jbar from l to l'. these transfers are only
c virtual. on exit bpn contains the corresponding new values of bp. f
c and d are not assumed symmetric. assert:k.le.n1-2
db1 = 0.d0
dbp2 = w1(ibar)*w2(jbar) + w3(ibar)*w4(jbar)
if (k.eq.0) go to 20
do 10 m=1,k
m1 = ls(m)
lm1 = ll(m)
f1 = f(ibar,m1)
f2 = f(m1,ibar)
d1 = d(jbar,lm1)
d2 = d(lm1,jbar)
db1 = db1 + f1*d1 + f2*d2
dbp2 = dbp2 - f2*w2(lm1) - f1*w4(lm1) - w1(m1)*d2 - w3(m1)*d1
10 continue
dbp2 = dbp2 + db1
20 continue
c the average objective function, obj, for the current partial
c permutation is calculable from the contents of the array bp,
c although we do not do so explicitly. we now construct delta, the
c anticipated change in obj, from the corresponding change in bp.
bpn(1) = bp(1) + db1
bpn(2) = bp(2) + dbp2
bpn(6) = bp(6) - w1(ibar) - w3(ibar)
bpn(7) = bp(7) - w2(jbar) - w4(jbar)
bpn(3) = bpn(6)*bpn(7)
if (k.ge.n1-2) bpn(3) = 0.d0
bpn(4) = bp(4) + c(ibar,jbar)
dbp5 = c(ibar,jbar)
kp1 = k + 1
c
do 30 m=kp1,n1
m1 = ls(m)
lm1 = ll(m)
dbp5 = dbp5 - c(ibar,lm1) - c(m1,jbar)
30 continue
c
bpn(5) = bp(5) + dbp5
delta = db1 + c(ibar,jbar) - rfnk*(bp(2)+bp(5)) +
* rfnk1*((bpn(3)*rfnk2-bp(3)*rfnk)+(bpn(2)+bpn(5)))
return
end
subroutine initrd(c,f,d,nc,nf,nd,pnorm,ls,ll,w1,w2,w3,w4,
+bp,bpn,bpb,rfnk,rfnk1,rfnk2,k,n1,n2,n3)
implicit doubleprecision (a-h,o-z)
dimension c(nc,1),f(nf,1),d(nd,1),ls(1),ll(1),w1(1),w2(1),
* w3(1), w4(1)
dimension bp(7),bpn(7),bpb(7)
c initialize recurrence data
c called by the quadratic assignment heuristic hgw.
scsl = 0.d0
sfs = 0.d0
sdl = 0.d0
cmx = 0.d0
fmx = 0.d0
dmx = 0.d0
rfnk = 1.0d0/dble(n1)
rfnk1 = 0.d0
rfnk1 = 1.0d0/dble(n1-1)
rfnk2 = 0.d0
if (n1.gt.2) rfnk2 = 1.0d0/dble(n1-2)
c
do 20 i=1,n1
ls(i) = i
ll(i) = i
c initialize row and col sums
fr = 0.d0
fc = 0.d0
dr = 0.d0
dc = 0.d0
c
do 10 j=1,n1
scsl = scsl + c(i,j)
cmx = max(cmx,abs(c(i,j)))
fr = fr + f(i,j)
fmx = max(fmx,abs(f(i,j)))
fc = fc + f(j,i)
dr = dr + d(i,j)
dmx = max(dmx,abs(d(i,j)))
dc = dc + d(j,i)
10 continue
c
sfs = sfs + fr
sdl = sdl + dr
w1(i) = fr
w3(i) = fc
w2(i) = dr
w4(i) = dc
20 continue
c
bp(1) = 0.d0
bp(2) = 0.d0
bp(3) = sfs*sdl
bp(4) = 0.d0
bp(5) = scsl
bp(6) = sfs
bp(7) = sdl
pnorm = cmx + fmx*dmx
c pnorm is an upper bound on the error in evaluating obj if n1 is
c less than the reciprocal of the machine precision
return
end
subroutine updrd(f,d,nf,nd,ls,ll,w1,w2,w3,w4,
+bp,bpn,bpb,rfnk,rfnk1,rfnk2,k,n1,n2,n3)
implicit doubleprecision (a-h,o-z)
dimension f(nf,1),d(nd,1),ls(1),ll(1),w1(1),w2(1),w3(1),w4(1)
dimension bp(7),bpn(7),bpb(7)
c update recurrence data just before k is incremented called by
c the quadratic assignment heuristic hgw. assert: k<=n1-2
kp1 = k + 1
kp11 = ls(kp1)
lkp1 = ll(kp1)
do 10 i=1,n1
w1(i) = w1(i) - f(i,kp11)
w3(i) = w3(i) - f(kp11,i)
w2(i) = w2(i) - d(i,lkp1)
w4(i) = w4(i) - d(lkp1,i)
10 continue
c
bp(1) = bpb(1)
bp(2) = bpb(2)
bp(3) = bpb(3)
bp(4) = bpb(4)
bp(5) = bpb(5)
bp(6) = bpb(6)
bp(7) = bpb(7)
rfnk = rfnk1
rfnk1 = rfnk2
rfnk2 = 0.d0
if (kp1.lt.n1-2) rfnk2 = 1.0d0/dble(n1-kp1-2)
c next value of k is kp1
return
end
doubleprecision function deltx(i, j, c, f, d, nc, nf, nd, loc3n,
+bp,bpn,bpb,rfnk,rfnk1,rfnk2,k,n1,n2,n3)
implicit doubleprecision (a-h,o-z)
dimension c(nc,1), f(nf,1), d(nd,1), loc3n(1)
dimension bp(7),bpn(7),bpb(7)
c called by the quadratic assignment heuristic hgw. returns the change
c (new - old) in the hgw objective function that would be produced if the
c existing assignment pairs (i<->li) and (j<->lj) were replaced by (i<->lj)
c and (j<->li). i and j are in s',li and lj are in l'. f and d are not
c assumed symmetric.
li = loc3n(i)
lj = loc3n(j)
deltx = c(i,lj) - c(i,li) + c(j,li) - c(j,lj) +
* (f(i,j)-f(j,i))*(d(lj,li)-d(li,lj))
c
do 10 m=1,n1
if (m.eq.i) go to 10
if (m.eq.j) go to 10
lm = loc3n(m)
deltx = deltx + (f(i,m)-f(j,m))*(d(lj,lm)-d(li,lm)) +
* (f(m,i)-f(m,j))*(d(lm,lj)-d(lm,li))
10 continue
c
return
end
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