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|
c--------------------------------------------------------------------------
c max clique algorithm based on quadratic zero-one programming
c This program is pardalos4 from Panos M. Pardalos & Gregory P. Rodgers
c described in "A branch and bound algorithm for the maximum clique
c problem", Operations Research, Vol. 19, No. 5 (1992), pp. 363-375.
c--------------------------------------------------------------------------
subroutine clique1(n,m,m2,np1,nwk,kat,hat,wcl,wk,x)
integer hat,kat,wk,wcl
logical x
dimension hat(m2),kat(np1),x(n),wk(nwk),wcl(n)
nprocs=1
do 1,i=1,nwk
wcl(i)=0
1 continue
c heuristic phase using the greedy method
call greedy(n,hat,kat,maxc,x,wk,wk(n+1),wk((2*n)+1),
> wk((3*n)+1) )
c branch and bound phase
ihsol= maxc
if (nprocs.ne.0)
>call bbnd(n,hat,kat,x,maxc,wk,nwk,nsubp,nsubpg,nprocs,nchg)
c heuristic size = ihsol ; maximum size = maxc
c number of min chages = nchg ; number of subproblems = nsubp
c minimizer subproblem = nsubpg
iptr = 0
do 10 i=1,n
wcl(i)=0
if (x(i)) then
iptr = iptr+1
wcl(iptr) = i
endif
10 continue
return
end
c= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
subroutine greedy(n,hat,kat,num1s,x,degfre,free,p,degfx1)
integer n,hat(*),kat(*),num1s,degfre(n),p(n),degfx1(n)
logical x(n),free(n),value
c
imax = 1
do 10 i=1,n
p(i) = i
degfre(i) = kat(i+1)-kat(i)
if (degfre(i).gt.degfre(imax)) imax= i
degfx1(i) = 0
free(i) = .true.
10 continue
c put the highest connected node into the greedy
num1s = 0
x(imax) = .true.
p(1) = imax
p(imax) = 1
free(imax) = .false.
c
lev = 1
20 continue
levp1 = lev+1
c --- update the vectors degfre, degfx1, and num1s
lp = p(lev)
c lp was the last variable that was fixed
if (x(lp)) then
num1s = num1s+1
do 70 k=kat(lp),kat(lp+1)-1
j = hat(k)
if (free(j)) then
degfre(j) = degfre(j) - 1
degfx1(j) = degfx1(j) + 1
endif
70 continue
else
c last vertex was thrown out so dont bother with degfx1
do 90 k=kat(lp),kat(lp+1)-1
degfre(hat(k)) = degfre(hat(k)) - 1
90 continue
endif
if (lev.eq.n) goto 1000
c
c --- determine if any variable can be fixed
do 130 i=levp1,n
ip = p(i)
c
c two tests to see if node ip can be fixed
c
c -1- is it disconnected to a node currently fixed to one?
if (degfx1(ip).lt.num1s) then
value = .false.
ivar = i
goto 150
endif
c
c -2- after the 1st test we know ip is connected to
c all nodes fixed to one. so now we can test if it
c is connected to all remaining free nodes. if it is
c then fix ip to 1
c mg if (degfre(ip).ge.n-levp) then
if (degfre(ip).ge.n-levp1) then
value = .true.
ivar = i
goto 150
endif
c
130 continue
c
c --- none could be forced so find variable with highest degree
ivar = levp1
do 140 i=levp1+1,n
if (degfre(p(i)).gt.degfre(p(ivar))) ivar=i
ivar = i
140 continue
c
c --- set to one and stack the subproblem with x(ip) = 0
value = .true.
c
c --- ivar is new fixed variable, increase lev & swap in p
150 lev = levp1
it = p(ivar)
p(ivar) = p(lev)
p(lev) = it
x(it) = value
free(it) = .false.
goto 20
1000 continue
return
c end greedy
end
c**************************************************************************
subroutine copyl(n,lfrom,lto)
logical lfrom(n),lto(n)
integer n
do 10 i=1,n
lto(i) = lfrom(i)
10 continue
return
c end of copyl
end
c**************************************************************************
subroutine bbnd(n, hat,kat,xmin,maxc,iwk,nwk,nsubp,nsubpg,nprocs,
> nchg)
integer n,nwk,hat(*),kat(*),maxc,iwk(nwk),nsubp,nsubpg,nprocs
logical xmin(n)
c local variables
integer csubp,cycles
integer nbusy,nfree,busy(6),free(6),strwk(6),split(6)
integer ldx,ldf,lstk,lp,lx,lfr,lxm,ltop,llev,lmin,lnsb,lnsg,lnwk,
> lend
integer nsplit,ipt,iptz,ipty,imin,imax,minspl,wsize,y,z,i,minsb
logical prempt,newmax
c --- initialize statistics
nchg = 0
nsubp = 0
nsubpg = 0
csubp = 0
cycles = 0
c --- initilize the mininum # of subproblems to solve in each cycle
minsb = 800
c --- divide workspace among nprocs processors
wsize = nwk/nprocs
strwk(1) = 0
do 10 i=2,nprocs
strwk(i) = strwk(i-1)+wsize
10 continue
c --- set up storage pointers
ldx = 1
ldf = ldx + ( n )
lstk= ldf + ( n )
lp = lstk + ( 3*(n+1))
lx = lp + ( n )
lfr = lx + ( n )
lxm = lfr + ( n )
ltop= lxm + ( n )
llev= ltop + ( 1 )
lmin= llev + ( 1 )
lnsb= lmin + ( 1 )
lnsg= lnsb + ( 1 )
lnwk= lnsg + ( 1 )
lend= lnwk + ( 1 )
if (lend.ge.wsize) then
c ww lend*nprocs,nwk
return
endif
c --- initialize subproblem 1
do 20 i=1,n
iwk(lp+i-1)= i
20 continue
iwk(ltop)= 1
iwk(lstk)= -1
iwk(llev)= 0
iwk(lnwk)= 0
iwk(lmin)= maxc
c --- set up busy and free lists
nbusy = 1
busy(1) = 1
nfree = nprocs-1
do 30 i=1,nfree
free(i) = i+1
30 continue
c-----------------------------------------------------------------------
c split subproblems till all procs are marked busy
c-----------------------------------------------------------------------
c while (nbusy.lt.nprocs.and.nbusy.gt.0)
1000 if (nbusy.lt.nprocs.and.nbusy.gt.0) then
c find all the split points in the nbusy subproblems
nsplit=0
do 40 i=1,nbusy
ipt = strwk(busy(i))
c is the top of the stack greater than 1?
if (iwk(ipt+ltop).gt.1) then
c get the level from the top element of the stack
split(i) = iwk(ipt + lstk + 3)
nsplit = nsplit + 1
else
split(i) = 0
endif
40 continue
c
if (nsplit.eq.0) then
c if there are no problems to split then force splitting
c of the first process in the busy list by (re)starting
c the branch and bound algorithm
50 ipt = strwk(busy(1))
call bbspl1(n,hat,kat,wsize,iwk(ipt+1),nchg)
nsubp = nsubp + iwk(ipt+lnsb)
csubp = csubp + iwk(ipt+lnsb)
c did bbspl1 find a new minimum ?
if (iwk(ipt+lmin).gt.maxc) then
maxc = iwk(ipt+lmin)
call copyl(n,iwk(ipt+lxm),xmin)
do 60 y=1,nprocs
iwk(strwk(y)+lmin) = maxc
60 continue
nsubpg = csubp + iwk(ipt+lnsg)
endif
c did bbspl1 solve the problem ?
if (iwk(ipt+llev).eq.-1) then
c attempted split solved the problem
c take y off busy list and add it to the free list
nfree = nfree + 1
nbusy = nbusy - 1
free(nfree) = busy(1)
do 70 i=1,nbusy
busy(i) = busy(i+1)
70 continue
c go back and try to split another subproblem
if (nbusy.gt.0) goto 50
else
c no, then a new split point was found
split(1) = iwk(ipt + lstk + 3)
nsplit = 1
endif
endif
c
if (nbusy.eq.0) goto 3000
c nsplit problems are now splitable
c find lowest splitpoint (highest pt in tree) and split
imin = 0
minspl = n+1
c examine splitpoints of all busy subproblems
do 80 i=1,nbusy
if (split(i).ne.0.and.split(i).lt.minspl) then
imin = i
minspl = split(imin)
endif
80 continue
c --- now split the process denoted by busy(imin)
y = busy(imin)
z = free(nfree)
ipty = strwk(y)
iptz = strwk(z)
c take z off free list and add it to the busy list
nfree = nfree - 1
nbusy = nbusy + 1
busy(nbusy) = z
c --- initialize subproblem z from subproblem y
c permutation vector p
do 90 i=1,n
iwk((iptz+lp-1)+i)= iwk((ipty+lp-1)+i)
90 continue
c initialize top of stack and stack as empty
iwk(iptz+ltop) = 1
iwk(iptz+lstk) = -1
iwk(iptz+lnwk) = 0
c level of execution
iwk(iptz+llev) = split(imin)
c incumbent
iwk(iptz+lmin)= maxc
c zero-one variables x
call copyl(n,iwk(ipty+lx),iwk(iptz+lx))
c change one x variable
call changl(iwk(iptz+lx),iwk(iptz+lp),split(imin))
c --> subroutine changl(x,p,lev)
c --> x(p(lev)) = .not.x(p(lev))
c remove subproblem from y's stack
call pllstk(iwk(ipty+ltop),iwk(ipty+lstk))
iwk(ipty+lnwk) = 0
c --> subroutine pllstk(topstk,stack)
c --> integer topstk,stack(3,*)
goto 1000
endif
c endwhile(nbusy.lt.nprocs.and.nbusy.gt.0)
c-----------------------------------------------------------------------
c solve subproblems in parallel
c-----------------------------------------------------------------------
c while (nbusy.eq.nprocs)
2000 if (nbusy.eq.nprocs) then
c execute nprocs in paralel this is called one cycle
c each proc will search its subproblem for maxvt vertices
cycles = cycles + 1
prempt = .false.
do 100 y=1,nprocs-1
ipt = strwk(y)+1
c solve subproblem y
call bbp(n,hat,kat,minsb,wsize,iwk(ipt),prempt,nchg)
cccc call dsptch('bbp',n,hat,kat,minsb,wsize,iwk(ipt),prempt)
100 continue
ipt = strwk(nprocs)+1
call bbp(n,hat,kat,minsb,wsize,iwk(ipt),prempt,nchg)
c
c synchronize subproblems (wait for all to finish)
cccc if (nprocs.gt.1) call syncro
c-----------------------------------------------------------------------
c analysis and synchronization phase
c-----------------------------------------------------------------------
nbusy = 0
nfree = 0
newmax = .false.
imax = 0
do 110 y=1,nprocs
ipt = strwk(y)
c iwk(ipt+lnsb)is the # of subproblems solved for this problem
nsubp = nsubp + iwk(ipt+lnsb)
c find which problem did the most to calculate critical path
if (iwk(ipt+lnsb).gt.imax) imax=iwk(ipt+lnsb)
if (iwk(ipt+lmin).gt.maxc) then
nsubpg = csubp + iwk(ipt+lnsg)
newmax = .true.
maxc = iwk(ipt+lmin)
call copyl(n,iwk(ipt+lxm),xmin)
endif
if (iwk(ipt+llev).ne.-1) then
nbusy = nbusy + 1
busy(nbusy) = y
else
nfree = nfree + 1
free(nfree) = y
endif
110 continue
c keep track of critical path count of subproblems
csubp = csubp + imax
c update new bound if one was found
if (newmax) then
do 120 y=1,nprocs
iwk(strwk(y)+lmin) = maxc
120 continue
endif
goto 2000
endif
c endwhile(nbusy.eq.nprocs)
c
c repeat until all subproblems have been solved (nbusy.eq.0)
if (nbusy.ne.0) goto 1000
c--------------------------------------------------------------------
c end of cyeturn to q01
c--------------------------------------------------------------------
3000 continue
c nb of cycles for solving the problem = cycles
c critical path subproblem count = csubp
200 format(' not enough workspace:',i8,' needed',i8,' specified')
return
c
c end of
end
c**************************************************************************
subroutine changl(x,p,lev)
c change the zero-one value in x(p(lev))
integer p(*),lev
logical x(*)
x(p(lev)) = .not.x(p(lev))
return
end
c**************************************************************************
subroutine pllstk(topstk,stack)
c pull the top most element off of the stack
integer topstk,stack(3,*)
topstk = topstk-1
c -1 is at top of stack. do not touch it
do 10 i=2,topstk
stack(1,i) = stack(1,i+1)
stack(2,i) = stack(2,i+1)
c cause bounds to be recalculated (save synchronization time)
stack(3,i) = -1
10 continue
return
end
c**************************************************************************
subroutine bbp(n,hat,kat,minsb,nwk,iwk,prempt,nchg)
integer n,hat(*),kat(*),minsb,nwk,iwk(nwk)
logical prempt
c local variables
integer maxwk
c --- set up storage pointers
ldx = 1
ldf = ldx + ( n )
lstk= ldf + ( n )
lp = lstk + ( 3*(n+1))
lx = lp + ( n )
lfr = lx + ( n )
lxm = lfr + ( n )
ltop= lxm + ( n )
llev= ltop + ( 1 )
lmin= llev + ( 1 )
lnsb= lmin + ( 1 )
lnsg= lnsb + ( 1 )
lnwk= lnsg + ( 1 )
lend= lnwk + ( 1 )
maxwk = nwk-lend
call bb(n,hat,kat,minsb,maxwk,
> iwk(ldx),iwk(ldf),iwk(lstk),iwk(lp),iwk(lx),
> iwk(lfr),iwk(lxm),
> iwk(ltop),iwk(llev),iwk(lmin),iwk(lnsb),iwk(lnsg),iwk(lnwk),
> iwk(lend),prempt,nchg)
return
c end of bbp
end
cprocess vector(report) directive('*vdir:')
c**************************************************************************
subroutine bb(n,hat,kat,maxs,maxwk,degfx1,degfre,stack,p,x,
> free,xmin,topstk,lev,maxc,nsubp,nsubpg,endwk,iwk,prempt,nchg)
c parameters
integer n,maxs,maxwk,topstk,lev,maxc,nsubp,nsubpg,endwk
integer hat(*),kat(*),degfx1(n),degfre(n),stack(3,n),p(n),
> iwk(maxwk)
logical x(n),xmin(n),free(n),prempt
c local variables
integer g,levp1,ip,lp,i,j,it,dgneed,num1s,nmlevp
integer nfree,newewk,itemp,iptr,ivar
logical uselst,value
c*vdir: prefer scalar on
nsubp = 0
nsubpg = 0
uselst = .false.
c calculate the upper bound on maxc, g
g = n
do 10 i=1,lev
ip = p(i)
free(ip) = .false.
if(.not.x(ip)) g = g-1
10 continue
do 20 i=lev+1,n
free(p(i)) = .true.
20 continue
30 if (lev.eq.-1.or.(nsubp.ge.maxs.and.prempt)) goto 1000
c
levp1 = lev + 1
c
if (g.le.maxc.or.lev.eq.n) then
c
c terminal node found
c
c --- check to see if it was a leaf
if (g.gt.maxc) then
maxc = g
do 40 j=1,n
xmin(j) = x(j)
40 continue
nsubpg = nsubp+1
nchg = nchg+1
endif
c
c --- get new subproblem from the stack
lev = stack(1,topstk)
g = stack(2,topstk)
itemp = stack(3,topstk)
topstk = topstk - 1
if (lev.ne.-1) then
x(p(lev)) = (.not.x(p(lev)))
do 50 i=lev+1,n
free(p(i)) = .true.
50 continue
c --- get degree counts from storage if there was enough room
if (itemp.ne.-1) then
endwk = itemp
itemp = endwk + (n-lev)
iptr = 0
do 55 i=1,n
if (free(i)) then
iptr = iptr+1
c the free list in p is now in order
p(lev+iptr) = i
degfre(i) = iwk(endwk+iptr)
degfx1(i) = iwk(itemp+iptr)
endif
55 continue
uselst = .true.
else
c there wasn't enough room, counts must be recalculated
uselst = .false.
endif
endif
nsubp = nsubp + 1
c
else
c
c proceed depth first
c
c --- calculate the degree of free nodes to free nodes
if (uselst) then
lp = p(lev)
c lp was the last variable that was fixed
if (x(lp)) then
c*vdir: prefer vector
c*vdir: ignore recrdeps(degfre,degfx1)
do 70 k=kat(lp),kat(lp+1)-1
degfre(hat(k)) = degfre(hat(k)) - 1
degfx1(hat(k)) = degfx1(hat(k)) + 1
70 continue
else
c last vertex was thrown out so dont bother with degfx1
c*vdir: prefer vector
c*vdir: ignore recrdeps(degfre)
do 90 k=kat(lp),kat(lp+1)-1
degfre(hat(k)) = degfre(hat(k)) - 1
90 continue
endif
else
c calculate free node connectivity from scratch
do 120 i = levp1,n
ip = p(i)
degfre(ip) = 0
degfx1(ip) = 0
do 110 k=kat(ip),kat(ip+1)-1
j = hat(k)
if (free(j)) then
degfre(ip) = degfre(ip) + 1
else
if (x(j)) degfx1(ip) = degfx1(ip) + 1
endif
110 continue
120 continue
endif
c
c --- determine if any variable can be fixed
num1s = lev + g - n
dgneed = maxc - num1s -1
nmlevp = n - levp1
do 130 i=levp1,n
ip = p(i)
c
c three tests to see if node ip can be fixed
c
c -1- is it disconnected to a node currently fixed to one?
if (degfx1(ip).lt.num1s) then
value = .false.
ivar = i
g = g - 1
goto 150
endif
c
c -2- after the 1st test we know ip is connected to
c all nodes fixed to one. so now we can test if it
c is connected to all remaining free nodes. if it is
c then fix ip to 1
if (degfre(ip).ge.nmlevp) then
value = .true.
ivar = i
goto 150
endif
c
c -3- is its total degree + 1 smaller than a known
c each free node needs at least degree dgneed to make
c a f size maxc. if not we can throw node away.
if (degfre(ip).le.dgneed) then
c fix to 0
value = .false.
g = g - 1
ivar = i
goto 150
endif
c
130 continue
c
c note: at this point we know that all free nodes are
c connected to all nodes fixed to one and that they
c are also disconnected to some free nodes. so
c we may choose variable to branch on based on how
c much they are connected or disconnected to other
c free nodes.
c
c --- none could be fixed so find variable i to branch on.
ccc rule1: choose node with highest free connectivity
c rule2: choose node with lowest free connectivity
c rule 2 is faster in verifying optimality and is used here
c rule 1 is the greedy method.
ivar = levp1
do 140 i=levp1+1,n
if (degfre(p(i)).lt.degfre(p(ivar))) ivar= i
140 continue
c
c --- set to one and stack the subproblem with x(ip) = 0
value = .true.
if (g-1.gt.maxc) then
c only stack subproblems that are not suboptimal
topstk = topstk + 1
stack(1,topstk) = levp1
stack(2,topstk) = g-1
nfree = n-levp1
newewk = endwk + (nfree+nfree)
if (newewk.le.maxwk) then
stack(3,topstk) = endwk
free(p(ivar)) = .false.
itemp = endwk + nfree
iptr = 0
do 145 i=1,n
c note: the following requires that the free list in
c the permutation p be in order
if (free(i)) then
iptr = iptr+1
iwk(endwk+iptr) = degfre(i)
iwk(itemp+iptr) = degfx1(i)
endif
145 continue
endwk = newewk
else
stack(3,topstk) = -1
endif
endif
c
c --- ivar is new fixed variable, increase lev & swap in p
150 lev = levp1
it = p(ivar)
p(ivar) = p(lev)
p(lev) = it
x(it) = value
free(it) = .false.
uselst = .true.
c
endif
goto 30
c
1000 prempt = .true.
return
c
c end of bb
end
c**************************************************************************
subroutine bbspl1(n,hat,kat,nwk,iwk,nchg)
integer n,hat(*),kat(*),nwk,iwk(nwk)
c --- set up storage pointers
ldx = 1
ldf = ldx + ( n )
lstk= ldf + ( n )
lp = lstk + ( 3*(n+1))
lx = lp + ( n )
lfr = lx + ( n )
lxm = lfr + ( n )
ltop= lxm + ( n )
llev= ltop + ( 1 )
lmin= llev + ( 1 )
lnsb= lmin + ( 1 )
lnsg= lnsb + ( 1 )
lnwk= lnsg + ( 1 )
lend= lnwk + ( 1 )
maxwk = nwk-lend
c bbspl2 will stop after the first subproblem is stacked
call bbspl2(n,hat,kat,maxwk,
> iwk(ldx),iwk(ldf),iwk(lstk),iwk(lp),iwk(lx),
> iwk(lfr),iwk(lxm),
> iwk(ltop),iwk(llev),iwk(lmin),iwk(lnsb),iwk(lnsg),iwk(lnwk),
> iwk(lend),nchg)
return
c
c end of bbspl1
end
c**************************************************************************
subroutine bbspl2(n,hat,kat,maxwk,
> degfx1,degfre,stack,p,x,free,xmin,
> topstk,lev,maxc,nsubp,nsubpg,endwk,iwk,nchg)
c parameters
integer n,maxwk,topstk,lev,maxc,nsubp,nsubpg,endwk
integer hat(*),kat(*),degfx1(n),degfre(n),stack(3,n),p(n),
> iwk(maxwk)
logical x(n),xmin(n),free(n)
c local variables
integer g,levp1,ip,lp,i,j,it,dgneed,num1s,nmlevp
integer nfree,newewk,itemp,iptr,ivar
logical uselst,value
c*vdir: prefer scalar on
nsubp = 0
nsubpg = 0
uselst = .false.
c calculate the upper bound on maxc, g
g = n
do 10 i=1,lev
ip = p(i)
free(ip) = .false.
if(.not.x(ip)) g = g-1
10 continue
do 20 i=lev+1,n
free(p(i)) = .true.
20 continue
30 if (lev.eq.-1.or.topstk.gt.1) goto 1000
c
levp1 = lev + 1
c
if (g.le.maxc.or.lev.eq.n) then
c
c terminal node found
c
c --- check to see if it was a leaf
if (g.gt.maxc) then
maxc = g
do 40 j=1,n
xmin(j) = x(j)
40 continue
nsubpg = nsubp+1
nchg = nchg + 1
endif
c
c --- get new subproblem from the stack
lev = stack(1,topstk)
g = stack(2,topstk)
itemp = stack(3,topstk)
topstk = topstk - 1
if (lev.ne.-1) then
x(p(lev)) = (.not.x(p(lev)))
do 50 i=lev+1,n
free(p(i)) = .true.
50 continue
c --- get degree counts from storage if there was enough room
if (itemp.ne.-1) then
endwk = itemp
itemp = endwk + (n-lev)
iptr = 0
do 55 i=1,n
if (free(i)) then
iptr = iptr+1
c the free list in p is now in order
p(lev+iptr) = i
degfre(i) = iwk(endwk+iptr)
degfx1(i) = iwk(itemp+iptr)
endif
55 continue
uselst = .true.
else
c there wasn't enough room, counts must be recalculated
uselst = .false.
endif
endif
nsubp = nsubp + 1
c
else
c
c proceed depth first
c
c --- calculate the degree of free nodes to free nodes
if (uselst) then
lp = p(lev)
c lp was the last variable that was fixed
if (x(lp)) then
c*vdir: prefer vector
c*vdir: ignore recrdeps(degfre,degfx1)
do 70 k=kat(lp),kat(lp+1)-1
degfre(hat(k)) = degfre(hat(k)) - 1
degfx1(hat(k)) = degfx1(hat(k)) + 1
70 continue
else
c last vertex was thrown out so dont bother with degfx1
c*vdir: prefer vector
c*vdir: ignore recrdeps(degfre)
do 90 k=kat(lp),kat(lp+1)-1
degfre(hat(k)) = degfre(hat(k)) - 1
90 continue
endif
else
c calculate free node connectivity from scratch
do 120 i = levp1,n
ip = p(i)
degfre(ip) = 0
degfx1(ip) = 0
do 110 k=kat(ip),kat(ip+1)-1
j = hat(k)
if (free(j)) then
degfre(ip) = degfre(ip) + 1
else
if (x(j)) degfx1(ip) = degfx1(ip) + 1
endif
110 continue
120 continue
endif
c
c --- determine if any variable can be fixed
num1s = lev + g - n
dgneed = maxc - num1s -1
nmlevp = n - levp1
do 130 i=levp1,n
ip = p(i)
c
c three tests to see if node ip can be fixed
c
c -1- is it disconnected to a node currently fixed to one?
if (degfx1(ip).lt.num1s) then
value = .false.
ivar = i
g = g - 1
goto 150
endif
c
c -2- after the 1st test we know ip is connected to
c all nodes fixed to one. so now we can test if it
c is connected to all remaining free nodes. if it is
c then fix ip to 1
if (degfre(ip).ge.nmlevp) then
value = .true.
ivar = i
goto 150
endif
c
c -3- is its total degree + 1 smaller than a known
c each free node needs at least degree dgneed to make
c a f size maxc. if not we can throw node away.
if (degfre(ip).le.dgneed) then
c fix to 0
value = .false.
g = g - 1
ivar = i
goto 150
endif
c
130 continue
c note: at this point we know that all free nodes are connected to all
c nodes connected to all nodes fixed to one and that they are also
c disconnected to some free nodes. so we may choose variable to branch on
c based on how much they are connected or disconnected to other free nodes.
c
c --- none could be fixed so find variable i to branch on.
ccc rule1: choose node with highest free connectivity
c rule2: choose node with lowest free connectivity
c rule 2 is faster in verifying optimality and is used here
c rule 1 is the greedy method.
ivar = levp1
do 140 i=levp1+1,n
if (degfre(p(i)).lt.degfre(p(ivar))) ivar=i
140 continue
c --- set to one and stack the subproblem with x(ip) = 0
value = .true.
if (g-1.gt.maxc) then
c only stack subproblems that are not suboptimal
topstk = topstk + 1
stack(1,topstk) = levp1
stack(2,topstk) = g-1
nfree = n-levp1
newewk = endwk + (nfree+nfree)
if (newewk.le.maxwk) then
stack(3,topstk) = endwk
free(p(ivar)) = .false.
itemp = endwk + nfree
iptr = 0
do 145 i=1,n
c note: the following requires that the free list in
c the permutation p be in order
if (free(i)) then
iptr = iptr+1
iwk(endwk+iptr) = degfre(i)
iwk(itemp+iptr) = degfx1(i)
endif
145 continue
endwk = newewk
else
stack(3,topstk) = -1
endif
endif
c --- ivar is new fixed variable, increase lev & swap in p
150 lev = levp1
it = p(ivar)
p(ivar) = p(lev)
p(lev) = it
x(it) = value
free(it) = .false.
uselst = .true.
c
endif
goto 30
c
1000 return
c end of bbspl2
end
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