Timur A. Fatkhullin
5279d1c41a
start developing of FITPACK C++ bindings mount_server.cpp: fix compilation error with GCC15
394 lines
13 KiB
Fortran
394 lines
13 KiB
Fortran
subroutine fppasu(iopt,ipar,idim,u,mu,v,mv,z,mz,s,nuest,nvest,
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* tol,maxit,nc,nu,tu,nv,tv,c,fp,fp0,fpold,reducu,reducv,fpintu,
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* fpintv,lastdi,nplusu,nplusv,nru,nrv,nrdatu,nrdatv,wrk,lwrk,ier)
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implicit none
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c ..
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c ..scalar arguments..
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real*8 s,tol,fp,fp0,fpold,reducu,reducv
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integer iopt,idim,mu,mv,mz,nuest,nvest,maxit,nc,nu,nv,lastdi,
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* nplusu,nplusv,lwrk,ier
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c ..array arguments..
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real*8 u(mu),v(mv),z(mz*idim),tu(nuest),tv(nvest),c(nc*idim),
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* fpintu(nuest),fpintv(nvest),wrk(lwrk)
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integer ipar(2),nrdatu(nuest),nrdatv(nvest),nru(mu),nrv(mv)
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c ..local scalars
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real*8 acc,fpms,f1,f2,f3,p,p1,p2,p3,rn,one,con1,con9,con4,
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* peru,perv,ub,ue,vb,ve
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integer i,ich1,ich3,ifbu,ifbv,ifsu,ifsv,iter,j,lau1,lav1,laa,
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* l,lau,lav,lbu,lbv,lq,lri,lsu,lsv,l1,l2,l3,l4,mm,mpm,mvnu,ncof,
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* nk1u,nk1v,nmaxu,nmaxv,nminu,nminv,nplu,nplv,npl1,nrintu,
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* nrintv,nue,nuk,nve,nuu,nvv
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c ..function references..
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real*8 abs,fprati
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integer max0,min0
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c ..subroutine references..
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c fpgrpa,fpknot
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c ..
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c set constants
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one = 1
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con1 = 0.1e0
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con9 = 0.9e0
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con4 = 0.4e-01
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c set boundaries of the approximation domain
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ub = u(1)
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ue = u(mu)
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vb = v(1)
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ve = v(mv)
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c we partition the working space.
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lsu = 1
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lsv = lsu+mu*4
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lri = lsv+mv*4
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mm = max0(nuest,mv)
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lq = lri+mm*idim
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mvnu = nuest*mv*idim
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lau = lq+mvnu
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nuk = nuest*5
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lbu = lau+nuk
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lav = lbu+nuk
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nuk = nvest*5
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lbv = lav+nuk
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laa = lbv+nuk
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lau1 = lau
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if(ipar(1).eq.0) go to 10
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peru = ue-ub
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lau1 = laa
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laa = laa+4*nuest
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10 lav1 = lav
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if(ipar(2).eq.0) go to 20
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perv = ve-vb
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lav1 = laa
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cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
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c part 1: determination of the number of knots and their position. c
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c **************************************************************** c
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c given a set of knots we compute the least-squares spline sinf(u,v), c
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c and the corresponding sum of squared residuals fp=f(p=inf). c
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c if iopt=-1 sinf(u,v) is the requested approximation. c
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c if iopt=0 or iopt=1 we check whether we can accept the knots: c
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c if fp <=s we will continue with the current set of knots. c
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c if fp > s we will increase the number of knots and compute the c
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c corresponding least-squares spline until finally fp<=s. c
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c the initial choice of knots depends on the value of s and iopt. c
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c if s=0 we have spline interpolation; in that case the number of c
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c knots equals nmaxu = mu+4+2*ipar(1) and nmaxv = mv+4+2*ipar(2) c
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c if s>0 and c
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c *iopt=0 we first compute the least-squares polynomial c
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c nu=nminu=8 and nv=nminv=8 c
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c *iopt=1 we start with the knots found at the last call of the c
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c routine, except for the case that s > fp0; then we can compute c
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c the least-squares polynomial directly. c
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cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
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c determine the number of knots for polynomial approximation.
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20 nminu = 8
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nminv = 8
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if(iopt.lt.0) go to 100
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c acc denotes the absolute tolerance for the root of f(p)=s.
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acc = tol*s
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c find nmaxu and nmaxv which denote the number of knots in u- and v-
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c direction in case of spline interpolation.
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nmaxu = mu+4+2*ipar(1)
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nmaxv = mv+4+2*ipar(2)
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c find nue and nve which denote the maximum number of knots
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c allowed in each direction
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nue = min0(nmaxu,nuest)
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nve = min0(nmaxv,nvest)
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if(s.gt.0.) go to 60
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c if s = 0, s(u,v) is an interpolating spline.
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nu = nmaxu
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nv = nmaxv
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c test whether the required storage space exceeds the available one.
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if(nv.gt.nvest .or. nu.gt.nuest) go to 420
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c find the position of the interior knots in case of interpolation.
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c the knots in the u-direction.
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nuu = nu-8
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if(nuu.eq.0) go to 40
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i = 5
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j = 3-ipar(1)
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do 30 l=1,nuu
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tu(i) = u(j)
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i = i+1
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j = j+1
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30 continue
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c the knots in the v-direction.
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40 nvv = nv-8
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if(nvv.eq.0) go to 60
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i = 5
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j = 3-ipar(2)
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do 50 l=1,nvv
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tv(i) = v(j)
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i = i+1
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j = j+1
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50 continue
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go to 100
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c if s > 0 our initial choice of knots depends on the value of iopt.
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60 if(iopt.eq.0) go to 90
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if(fp0.le.s) go to 90
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c if iopt=1 and fp0 > s we start computing the least- squares spline
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c according to the set of knots found at the last call of the routine.
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c we determine the number of grid coordinates u(i) inside each knot
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c interval (tu(l),tu(l+1)).
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l = 5
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j = 1
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nrdatu(1) = 0
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mpm = mu-1
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do 70 i=2,mpm
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nrdatu(j) = nrdatu(j)+1
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if(u(i).lt.tu(l)) go to 70
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nrdatu(j) = nrdatu(j)-1
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l = l+1
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j = j+1
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nrdatu(j) = 0
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70 continue
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c we determine the number of grid coordinates v(i) inside each knot
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c interval (tv(l),tv(l+1)).
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l = 5
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j = 1
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nrdatv(1) = 0
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mpm = mv-1
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do 80 i=2,mpm
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nrdatv(j) = nrdatv(j)+1
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if(v(i).lt.tv(l)) go to 80
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nrdatv(j) = nrdatv(j)-1
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l = l+1
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j = j+1
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nrdatv(j) = 0
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80 continue
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go to 100
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c if iopt=0 or iopt=1 and s>=fp0, we start computing the least-squares
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c polynomial (which is a spline without interior knots).
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90 nu = nminu
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nv = nminv
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nrdatu(1) = mu-2
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nrdatv(1) = mv-2
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lastdi = 0
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nplusu = 0
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nplusv = 0
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fp0 = 0.
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fpold = 0.
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reducu = 0.
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reducv = 0.
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100 mpm = mu+mv
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ifsu = 0
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ifsv = 0
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ifbu = 0
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ifbv = 0
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p = -one
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c main loop for the different sets of knots.mpm=mu+mv is a save upper
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c bound for the number of trials.
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do 250 iter=1,mpm
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if(nu.eq.nminu .and. nv.eq.nminv) ier = -2
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c find nrintu (nrintv) which is the number of knot intervals in the
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c u-direction (v-direction).
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nrintu = nu-nminu+1
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nrintv = nv-nminv+1
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c find ncof, the number of b-spline coefficients for the current set
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c of knots.
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nk1u = nu-4
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nk1v = nv-4
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ncof = nk1u*nk1v
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c find the position of the additional knots which are needed for the
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c b-spline representation of s(u,v).
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if(ipar(1).ne.0) go to 110
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i = nu
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do 105 j=1,4
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tu(j) = ub
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tu(i) = ue
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i = i-1
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105 continue
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go to 120
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110 l1 = 4
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l2 = l1
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l3 = nu-3
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l4 = l3
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tu(l2) = ub
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tu(l3) = ue
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do 115 j=1,3
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l1 = l1+1
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l2 = l2-1
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l3 = l3+1
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l4 = l4-1
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tu(l2) = tu(l4)-peru
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tu(l3) = tu(l1)+peru
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115 continue
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120 if(ipar(2).ne.0) go to 130
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i = nv
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do 125 j=1,4
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tv(j) = vb
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tv(i) = ve
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i = i-1
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125 continue
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go to 140
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130 l1 = 4
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l2 = l1
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l3 = nv-3
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l4 = l3
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tv(l2) = vb
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tv(l3) = ve
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do 135 j=1,3
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l1 = l1+1
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l2 = l2-1
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l3 = l3+1
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l4 = l4-1
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tv(l2) = tv(l4)-perv
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tv(l3) = tv(l1)+perv
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135 continue
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c find the least-squares spline sinf(u,v) and calculate for each knot
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c interval tu(j+3)<=u<=tu(j+4) (tv(j+3)<=v<=tv(j+4)) the sum
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c of squared residuals fpintu(j),j=1,2,...,nu-7 (fpintv(j),j=1,2,...
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c ,nv-7) for the data points having their absciss (ordinate)-value
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c belonging to that interval.
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c fp gives the total sum of squared residuals.
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140 call fpgrpa(ifsu,ifsv,ifbu,ifbv,idim,ipar,u,mu,v,mv,z,mz,tu,
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* nu,tv,nv,p,c,nc,fp,fpintu,fpintv,mm,mvnu,wrk(lsu),wrk(lsv),
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* wrk(lri),wrk(lq),wrk(lau),wrk(lau1),wrk(lav),wrk(lav1),
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* wrk(lbu),wrk(lbv),nru,nrv)
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if(ier.eq.(-2)) fp0 = fp
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c test whether the least-squares spline is an acceptable solution.
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if(iopt.lt.0) go to 440
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fpms = fp-s
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if(abs(fpms) .lt. acc) go to 440
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c if f(p=inf) < s, we accept the choice of knots.
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if(fpms.lt.0.) go to 300
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c if nu=nmaxu and nv=nmaxv, sinf(u,v) is an interpolating spline.
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if(nu.eq.nmaxu .and. nv.eq.nmaxv) go to 430
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c increase the number of knots.
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c if nu=nue and nv=nve we cannot further increase the number of knots
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c because of the storage capacity limitation.
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if(nu.eq.nue .and. nv.eq.nve) go to 420
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ier = 0
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c adjust the parameter reducu or reducv according to the direction
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c in which the last added knots were located.
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if (lastdi.lt.0) go to 150
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if (lastdi.eq.0) go to 170
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go to 160
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150 reducu = fpold-fp
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go to 170
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160 reducv = fpold-fp
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c store the sum of squared residuals for the current set of knots.
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170 fpold = fp
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c find nplu, the number of knots we should add in the u-direction.
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nplu = 1
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if(nu.eq.nminu) go to 180
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npl1 = nplusu*2
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rn = nplusu
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if(reducu.gt.acc) npl1 = rn*fpms/reducu
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nplu = min0(nplusu*2,max0(npl1,nplusu/2,1))
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c find nplv, the number of knots we should add in the v-direction.
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180 nplv = 1
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if(nv.eq.nminv) go to 190
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npl1 = nplusv*2
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rn = nplusv
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if(reducv.gt.acc) npl1 = rn*fpms/reducv
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nplv = min0(nplusv*2,max0(npl1,nplusv/2,1))
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190 if (nplu.lt.nplv) go to 210
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if (nplu.eq.nplv) go to 200
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go to 230
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200 if(lastdi.lt.0) go to 230
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210 if(nu.eq.nue) go to 230
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c addition in the u-direction.
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lastdi = -1
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nplusu = nplu
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ifsu = 0
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do 220 l=1,nplusu
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c add a new knot in the u-direction
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call fpknot(u,mu,tu,nu,fpintu,nrdatu,nrintu,nuest,1)
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c test whether we cannot further increase the number of knots in the
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c u-direction.
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if(nu.eq.nue) go to 250
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220 continue
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go to 250
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230 if(nv.eq.nve) go to 210
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c addition in the v-direction.
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lastdi = 1
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nplusv = nplv
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ifsv = 0
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do 240 l=1,nplusv
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c add a new knot in the v-direction.
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call fpknot(v,mv,tv,nv,fpintv,nrdatv,nrintv,nvest,1)
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c test whether we cannot further increase the number of knots in the
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c v-direction.
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if(nv.eq.nve) go to 250
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240 continue
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c restart the computations with the new set of knots.
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250 continue
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c test whether the least-squares polynomial is a solution of our
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c approximation problem.
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300 if(ier.eq.(-2)) go to 440
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cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
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c part 2: determination of the smoothing spline sp(u,v) c
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c ***************************************************** c
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c we have determined the number of knots and their position. we now c
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c compute the b-spline coefficients of the smoothing spline sp(u,v). c
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c this smoothing spline varies with the parameter p in such a way thatc
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c f(p)=suml=1,idim(sumi=1,mu(sumj=1,mv((z(i,j,l)-sp(u(i),v(j),l))**2) c
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c is a continuous, strictly decreasing function of p. moreover the c
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c least-squares polynomial corresponds to p=0 and the least-squares c
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c spline to p=infinity. iteratively we then have to determine the c
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c positive value of p such that f(p)=s. the process which is proposed c
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c here makes use of rational interpolation. f(p) is approximated by a c
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c rational function r(p)=(u*p+v)/(p+w); three values of p (p1,p2,p3) c
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c with corresponding values of f(p) (f1=f(p1)-s,f2=f(p2)-s,f3=f(p3)-s)c
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c are used to calculate the new value of p such that r(p)=s. c
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c convergence is guaranteed by taking f1 > 0 and f3 < 0. c
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cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
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c initial value for p.
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p1 = 0.
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f1 = fp0-s
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p3 = -one
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f3 = fpms
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p = one
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ich1 = 0
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ich3 = 0
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c iteration process to find the root of f(p)=s.
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do 350 iter = 1,maxit
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c find the smoothing spline sp(u,v) and the corresponding sum of
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c squared residuals fp.
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call fpgrpa(ifsu,ifsv,ifbu,ifbv,idim,ipar,u,mu,v,mv,z,mz,tu,
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* nu,tv,nv,p,c,nc,fp,fpintu,fpintv,mm,mvnu,wrk(lsu),wrk(lsv),
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* wrk(lri),wrk(lq),wrk(lau),wrk(lau1),wrk(lav),wrk(lav1),
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* wrk(lbu),wrk(lbv),nru,nrv)
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c test whether the approximation sp(u,v) is an acceptable solution.
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fpms = fp-s
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if(abs(fpms).lt.acc) go to 440
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c test whether the maximum allowable number of iterations has been
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c reached.
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if(iter.eq.maxit) go to 400
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c carry out one more step of the iteration process.
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p2 = p
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f2 = fpms
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if(ich3.ne.0) go to 320
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if((f2-f3).gt.acc) go to 310
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c our initial choice of p is too large.
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p3 = p2
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f3 = f2
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p = p*con4
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if(p.le.p1) p = p1*con9 + p2*con1
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go to 350
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310 if(f2.lt.0.) ich3 = 1
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320 if(ich1.ne.0) go to 340
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if((f1-f2).gt.acc) go to 330
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c our initial choice of p is too small
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p1 = p2
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f1 = f2
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p = p/con4
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if(p3.lt.0.) go to 350
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if(p.ge.p3) p = p2*con1 + p3*con9
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go to 350
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c test whether the iteration process proceeds as theoretically
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c expected.
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330 if(f2.gt.0.) ich1 = 1
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340 if(f2.ge.f1 .or. f2.le.f3) go to 410
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c find the new value of p.
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p = fprati(p1,f1,p2,f2,p3,f3)
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350 continue
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c error codes and messages.
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400 ier = 3
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go to 440
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410 ier = 2
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go to 440
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420 ier = 1
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go to 440
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430 ier = -1
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fp = 0.
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440 return
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end
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