394 lines
13 KiB
Fortran
394 lines
13 KiB
Fortran
subroutine fppasu(iopt,ipar,idim,u,mu,v,mv,z,mz,s,nuest,nvest,
|
|
* tol,maxit,nc,nu,tu,nv,tv,c,fp,fp0,fpold,reducu,reducv,fpintu,
|
|
* fpintv,lastdi,nplusu,nplusv,nru,nrv,nrdatu,nrdatv,wrk,lwrk,ier)
|
|
implicit none
|
|
c ..
|
|
c ..scalar arguments..
|
|
real*8 s,tol,fp,fp0,fpold,reducu,reducv
|
|
integer iopt,idim,mu,mv,mz,nuest,nvest,maxit,nc,nu,nv,lastdi,
|
|
* nplusu,nplusv,lwrk,ier
|
|
c ..array arguments..
|
|
real*8 u(mu),v(mv),z(mz*idim),tu(nuest),tv(nvest),c(nc*idim),
|
|
* fpintu(nuest),fpintv(nvest),wrk(lwrk)
|
|
integer ipar(2),nrdatu(nuest),nrdatv(nvest),nru(mu),nrv(mv)
|
|
c ..local scalars
|
|
real*8 acc,fpms,f1,f2,f3,p,p1,p2,p3,rn,one,con1,con9,con4,
|
|
* peru,perv,ub,ue,vb,ve
|
|
integer i,ich1,ich3,ifbu,ifbv,ifsu,ifsv,iter,j,lau1,lav1,laa,
|
|
* l,lau,lav,lbu,lbv,lq,lri,lsu,lsv,l1,l2,l3,l4,mm,mpm,mvnu,ncof,
|
|
* nk1u,nk1v,nmaxu,nmaxv,nminu,nminv,nplu,nplv,npl1,nrintu,
|
|
* nrintv,nue,nuk,nve,nuu,nvv
|
|
c ..function references..
|
|
real*8 abs,fprati
|
|
integer max0,min0
|
|
c ..subroutine references..
|
|
c fpgrpa,fpknot
|
|
c ..
|
|
c set constants
|
|
one = 1
|
|
con1 = 0.1e0
|
|
con9 = 0.9e0
|
|
con4 = 0.4e-01
|
|
c set boundaries of the approximation domain
|
|
ub = u(1)
|
|
ue = u(mu)
|
|
vb = v(1)
|
|
ve = v(mv)
|
|
c we partition the working space.
|
|
lsu = 1
|
|
lsv = lsu+mu*4
|
|
lri = lsv+mv*4
|
|
mm = max0(nuest,mv)
|
|
lq = lri+mm*idim
|
|
mvnu = nuest*mv*idim
|
|
lau = lq+mvnu
|
|
nuk = nuest*5
|
|
lbu = lau+nuk
|
|
lav = lbu+nuk
|
|
nuk = nvest*5
|
|
lbv = lav+nuk
|
|
laa = lbv+nuk
|
|
lau1 = lau
|
|
if(ipar(1).eq.0) go to 10
|
|
peru = ue-ub
|
|
lau1 = laa
|
|
laa = laa+4*nuest
|
|
10 lav1 = lav
|
|
if(ipar(2).eq.0) go to 20
|
|
perv = ve-vb
|
|
lav1 = laa
|
|
cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
|
|
c part 1: determination of the number of knots and their position. c
|
|
c **************************************************************** c
|
|
c given a set of knots we compute the least-squares spline sinf(u,v), c
|
|
c and the corresponding sum of squared residuals fp=f(p=inf). c
|
|
c if iopt=-1 sinf(u,v) is the requested approximation. c
|
|
c if iopt=0 or iopt=1 we check whether we can accept the knots: c
|
|
c if fp <=s we will continue with the current set of knots. c
|
|
c if fp > s we will increase the number of knots and compute the c
|
|
c corresponding least-squares spline until finally fp<=s. c
|
|
c the initial choice of knots depends on the value of s and iopt. c
|
|
c if s=0 we have spline interpolation; in that case the number of c
|
|
c knots equals nmaxu = mu+4+2*ipar(1) and nmaxv = mv+4+2*ipar(2) c
|
|
c if s>0 and c
|
|
c *iopt=0 we first compute the least-squares polynomial c
|
|
c nu=nminu=8 and nv=nminv=8 c
|
|
c *iopt=1 we start with the knots found at the last call of the c
|
|
c routine, except for the case that s > fp0; then we can compute c
|
|
c the least-squares polynomial directly. c
|
|
cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
|
|
c determine the number of knots for polynomial approximation.
|
|
20 nminu = 8
|
|
nminv = 8
|
|
if(iopt.lt.0) go to 100
|
|
c acc denotes the absolute tolerance for the root of f(p)=s.
|
|
acc = tol*s
|
|
c find nmaxu and nmaxv which denote the number of knots in u- and v-
|
|
c direction in case of spline interpolation.
|
|
nmaxu = mu+4+2*ipar(1)
|
|
nmaxv = mv+4+2*ipar(2)
|
|
c find nue and nve which denote the maximum number of knots
|
|
c allowed in each direction
|
|
nue = min0(nmaxu,nuest)
|
|
nve = min0(nmaxv,nvest)
|
|
if(s.gt.0.) go to 60
|
|
c if s = 0, s(u,v) is an interpolating spline.
|
|
nu = nmaxu
|
|
nv = nmaxv
|
|
c test whether the required storage space exceeds the available one.
|
|
if(nv.gt.nvest .or. nu.gt.nuest) go to 420
|
|
c find the position of the interior knots in case of interpolation.
|
|
c the knots in the u-direction.
|
|
nuu = nu-8
|
|
if(nuu.eq.0) go to 40
|
|
i = 5
|
|
j = 3-ipar(1)
|
|
do 30 l=1,nuu
|
|
tu(i) = u(j)
|
|
i = i+1
|
|
j = j+1
|
|
30 continue
|
|
c the knots in the v-direction.
|
|
40 nvv = nv-8
|
|
if(nvv.eq.0) go to 60
|
|
i = 5
|
|
j = 3-ipar(2)
|
|
do 50 l=1,nvv
|
|
tv(i) = v(j)
|
|
i = i+1
|
|
j = j+1
|
|
50 continue
|
|
go to 100
|
|
c if s > 0 our initial choice of knots depends on the value of iopt.
|
|
60 if(iopt.eq.0) go to 90
|
|
if(fp0.le.s) go to 90
|
|
c if iopt=1 and fp0 > s we start computing the least- squares spline
|
|
c according to the set of knots found at the last call of the routine.
|
|
c we determine the number of grid coordinates u(i) inside each knot
|
|
c interval (tu(l),tu(l+1)).
|
|
l = 5
|
|
j = 1
|
|
nrdatu(1) = 0
|
|
mpm = mu-1
|
|
do 70 i=2,mpm
|
|
nrdatu(j) = nrdatu(j)+1
|
|
if(u(i).lt.tu(l)) go to 70
|
|
nrdatu(j) = nrdatu(j)-1
|
|
l = l+1
|
|
j = j+1
|
|
nrdatu(j) = 0
|
|
70 continue
|
|
c we determine the number of grid coordinates v(i) inside each knot
|
|
c interval (tv(l),tv(l+1)).
|
|
l = 5
|
|
j = 1
|
|
nrdatv(1) = 0
|
|
mpm = mv-1
|
|
do 80 i=2,mpm
|
|
nrdatv(j) = nrdatv(j)+1
|
|
if(v(i).lt.tv(l)) go to 80
|
|
nrdatv(j) = nrdatv(j)-1
|
|
l = l+1
|
|
j = j+1
|
|
nrdatv(j) = 0
|
|
80 continue
|
|
go to 100
|
|
c if iopt=0 or iopt=1 and s>=fp0, we start computing the least-squares
|
|
c polynomial (which is a spline without interior knots).
|
|
90 nu = nminu
|
|
nv = nminv
|
|
nrdatu(1) = mu-2
|
|
nrdatv(1) = mv-2
|
|
lastdi = 0
|
|
nplusu = 0
|
|
nplusv = 0
|
|
fp0 = 0.
|
|
fpold = 0.
|
|
reducu = 0.
|
|
reducv = 0.
|
|
100 mpm = mu+mv
|
|
ifsu = 0
|
|
ifsv = 0
|
|
ifbu = 0
|
|
ifbv = 0
|
|
p = -one
|
|
c main loop for the different sets of knots.mpm=mu+mv is a save upper
|
|
c bound for the number of trials.
|
|
do 250 iter=1,mpm
|
|
if(nu.eq.nminu .and. nv.eq.nminv) ier = -2
|
|
c find nrintu (nrintv) which is the number of knot intervals in the
|
|
c u-direction (v-direction).
|
|
nrintu = nu-nminu+1
|
|
nrintv = nv-nminv+1
|
|
c find ncof, the number of b-spline coefficients for the current set
|
|
c of knots.
|
|
nk1u = nu-4
|
|
nk1v = nv-4
|
|
ncof = nk1u*nk1v
|
|
c find the position of the additional knots which are needed for the
|
|
c b-spline representation of s(u,v).
|
|
if(ipar(1).ne.0) go to 110
|
|
i = nu
|
|
do 105 j=1,4
|
|
tu(j) = ub
|
|
tu(i) = ue
|
|
i = i-1
|
|
105 continue
|
|
go to 120
|
|
110 l1 = 4
|
|
l2 = l1
|
|
l3 = nu-3
|
|
l4 = l3
|
|
tu(l2) = ub
|
|
tu(l3) = ue
|
|
do 115 j=1,3
|
|
l1 = l1+1
|
|
l2 = l2-1
|
|
l3 = l3+1
|
|
l4 = l4-1
|
|
tu(l2) = tu(l4)-peru
|
|
tu(l3) = tu(l1)+peru
|
|
115 continue
|
|
120 if(ipar(2).ne.0) go to 130
|
|
i = nv
|
|
do 125 j=1,4
|
|
tv(j) = vb
|
|
tv(i) = ve
|
|
i = i-1
|
|
125 continue
|
|
go to 140
|
|
130 l1 = 4
|
|
l2 = l1
|
|
l3 = nv-3
|
|
l4 = l3
|
|
tv(l2) = vb
|
|
tv(l3) = ve
|
|
do 135 j=1,3
|
|
l1 = l1+1
|
|
l2 = l2-1
|
|
l3 = l3+1
|
|
l4 = l4-1
|
|
tv(l2) = tv(l4)-perv
|
|
tv(l3) = tv(l1)+perv
|
|
135 continue
|
|
c find the least-squares spline sinf(u,v) and calculate for each knot
|
|
c interval tu(j+3)<=u<=tu(j+4) (tv(j+3)<=v<=tv(j+4)) the sum
|
|
c of squared residuals fpintu(j),j=1,2,...,nu-7 (fpintv(j),j=1,2,...
|
|
c ,nv-7) for the data points having their absciss (ordinate)-value
|
|
c belonging to that interval.
|
|
c fp gives the total sum of squared residuals.
|
|
140 call fpgrpa(ifsu,ifsv,ifbu,ifbv,idim,ipar,u,mu,v,mv,z,mz,tu,
|
|
* nu,tv,nv,p,c,nc,fp,fpintu,fpintv,mm,mvnu,wrk(lsu),wrk(lsv),
|
|
* wrk(lri),wrk(lq),wrk(lau),wrk(lau1),wrk(lav),wrk(lav1),
|
|
* wrk(lbu),wrk(lbv),nru,nrv)
|
|
if(ier.eq.(-2)) fp0 = fp
|
|
c test whether the least-squares spline is an acceptable solution.
|
|
if(iopt.lt.0) go to 440
|
|
fpms = fp-s
|
|
if(abs(fpms) .lt. acc) go to 440
|
|
c if f(p=inf) < s, we accept the choice of knots.
|
|
if(fpms.lt.0.) go to 300
|
|
c if nu=nmaxu and nv=nmaxv, sinf(u,v) is an interpolating spline.
|
|
if(nu.eq.nmaxu .and. nv.eq.nmaxv) go to 430
|
|
c increase the number of knots.
|
|
c if nu=nue and nv=nve we cannot further increase the number of knots
|
|
c because of the storage capacity limitation.
|
|
if(nu.eq.nue .and. nv.eq.nve) go to 420
|
|
ier = 0
|
|
c adjust the parameter reducu or reducv according to the direction
|
|
c in which the last added knots were located.
|
|
if (lastdi.lt.0) go to 150
|
|
if (lastdi.eq.0) go to 170
|
|
go to 160
|
|
150 reducu = fpold-fp
|
|
go to 170
|
|
160 reducv = fpold-fp
|
|
c store the sum of squared residuals for the current set of knots.
|
|
170 fpold = fp
|
|
c find nplu, the number of knots we should add in the u-direction.
|
|
nplu = 1
|
|
if(nu.eq.nminu) go to 180
|
|
npl1 = nplusu*2
|
|
rn = nplusu
|
|
if(reducu.gt.acc) npl1 = rn*fpms/reducu
|
|
nplu = min0(nplusu*2,max0(npl1,nplusu/2,1))
|
|
c find nplv, the number of knots we should add in the v-direction.
|
|
180 nplv = 1
|
|
if(nv.eq.nminv) go to 190
|
|
npl1 = nplusv*2
|
|
rn = nplusv
|
|
if(reducv.gt.acc) npl1 = rn*fpms/reducv
|
|
nplv = min0(nplusv*2,max0(npl1,nplusv/2,1))
|
|
190 if (nplu.lt.nplv) go to 210
|
|
if (nplu.eq.nplv) go to 200
|
|
go to 230
|
|
200 if(lastdi.lt.0) go to 230
|
|
210 if(nu.eq.nue) go to 230
|
|
c addition in the u-direction.
|
|
lastdi = -1
|
|
nplusu = nplu
|
|
ifsu = 0
|
|
do 220 l=1,nplusu
|
|
c add a new knot in the u-direction
|
|
call fpknot(u,mu,tu,nu,fpintu,nrdatu,nrintu,nuest,1)
|
|
c test whether we cannot further increase the number of knots in the
|
|
c u-direction.
|
|
if(nu.eq.nue) go to 250
|
|
220 continue
|
|
go to 250
|
|
230 if(nv.eq.nve) go to 210
|
|
c addition in the v-direction.
|
|
lastdi = 1
|
|
nplusv = nplv
|
|
ifsv = 0
|
|
do 240 l=1,nplusv
|
|
c add a new knot in the v-direction.
|
|
call fpknot(v,mv,tv,nv,fpintv,nrdatv,nrintv,nvest,1)
|
|
c test whether we cannot further increase the number of knots in the
|
|
c v-direction.
|
|
if(nv.eq.nve) go to 250
|
|
240 continue
|
|
c restart the computations with the new set of knots.
|
|
250 continue
|
|
c test whether the least-squares polynomial is a solution of our
|
|
c approximation problem.
|
|
300 if(ier.eq.(-2)) go to 440
|
|
cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
|
|
c part 2: determination of the smoothing spline sp(u,v) c
|
|
c ***************************************************** c
|
|
c we have determined the number of knots and their position. we now c
|
|
c compute the b-spline coefficients of the smoothing spline sp(u,v). c
|
|
c this smoothing spline varies with the parameter p in such a way thatc
|
|
c f(p)=suml=1,idim(sumi=1,mu(sumj=1,mv((z(i,j,l)-sp(u(i),v(j),l))**2) c
|
|
c is a continuous, strictly decreasing function of p. moreover the c
|
|
c least-squares polynomial corresponds to p=0 and the least-squares c
|
|
c spline to p=infinity. iteratively we then have to determine the c
|
|
c positive value of p such that f(p)=s. the process which is proposed c
|
|
c here makes use of rational interpolation. f(p) is approximated by a c
|
|
c rational function r(p)=(u*p+v)/(p+w); three values of p (p1,p2,p3) c
|
|
c with corresponding values of f(p) (f1=f(p1)-s,f2=f(p2)-s,f3=f(p3)-s)c
|
|
c are used to calculate the new value of p such that r(p)=s. c
|
|
c convergence is guaranteed by taking f1 > 0 and f3 < 0. c
|
|
cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
|
|
c initial value for p.
|
|
p1 = 0.
|
|
f1 = fp0-s
|
|
p3 = -one
|
|
f3 = fpms
|
|
p = one
|
|
ich1 = 0
|
|
ich3 = 0
|
|
c iteration process to find the root of f(p)=s.
|
|
do 350 iter = 1,maxit
|
|
c find the smoothing spline sp(u,v) and the corresponding sum of
|
|
c squared residuals fp.
|
|
call fpgrpa(ifsu,ifsv,ifbu,ifbv,idim,ipar,u,mu,v,mv,z,mz,tu,
|
|
* nu,tv,nv,p,c,nc,fp,fpintu,fpintv,mm,mvnu,wrk(lsu),wrk(lsv),
|
|
* wrk(lri),wrk(lq),wrk(lau),wrk(lau1),wrk(lav),wrk(lav1),
|
|
* wrk(lbu),wrk(lbv),nru,nrv)
|
|
c test whether the approximation sp(u,v) is an acceptable solution.
|
|
fpms = fp-s
|
|
if(abs(fpms).lt.acc) go to 440
|
|
c test whether the maximum allowable number of iterations has been
|
|
c reached.
|
|
if(iter.eq.maxit) go to 400
|
|
c carry out one more step of the iteration process.
|
|
p2 = p
|
|
f2 = fpms
|
|
if(ich3.ne.0) go to 320
|
|
if((f2-f3).gt.acc) go to 310
|
|
c our initial choice of p is too large.
|
|
p3 = p2
|
|
f3 = f2
|
|
p = p*con4
|
|
if(p.le.p1) p = p1*con9 + p2*con1
|
|
go to 350
|
|
310 if(f2.lt.0.) ich3 = 1
|
|
320 if(ich1.ne.0) go to 340
|
|
if((f1-f2).gt.acc) go to 330
|
|
c our initial choice of p is too small
|
|
p1 = p2
|
|
f1 = f2
|
|
p = p/con4
|
|
if(p3.lt.0.) go to 350
|
|
if(p.ge.p3) p = p2*con1 + p3*con9
|
|
go to 350
|
|
c test whether the iteration process proceeds as theoretically
|
|
c expected.
|
|
330 if(f2.gt.0.) ich1 = 1
|
|
340 if(f2.ge.f1 .or. f2.le.f3) go to 410
|
|
c find the new value of p.
|
|
p = fprati(p1,f1,p2,f2,p3,f3)
|
|
350 continue
|
|
c error codes and messages.
|
|
400 ier = 3
|
|
go to 440
|
|
410 ier = 2
|
|
go to 440
|
|
420 ier = 1
|
|
go to 440
|
|
430 ier = -1
|
|
fp = 0.
|
|
440 return
|
|
end
|