Timur A. Fatkhullin
5279d1c41a
start developing of FITPACK C++ bindings mount_server.cpp: fix compilation error with GCC15
766 lines
25 KiB
Fortran
766 lines
25 KiB
Fortran
recursive subroutine fpsphe(iopt,m,teta,phi,r,w,s,ntest,npest,
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* eta,tol,maxit,
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* ib1,ib3,nc,ncc,intest,nrest,nt,tt,np,tp,c,fp,sup,fpint,coord,f,
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* ff,row,coco,cosi,a,q,bt,bp,spt,spp,h,index,nummer,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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integer iopt,m,ntest,npest,maxit,ib1,ib3,nc,ncc,intest,nrest,
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* nt,np,lwrk,ier
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real*8 s,eta,tol,fp,sup
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c ..array arguments..
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real*8 teta(m),phi(m),r(m),w(m),tt(ntest),tp(npest),c(nc),
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* fpint(intest),coord(intest),f(ncc),ff(nc),row(npest),coco(npest),
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*
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* cosi(npest),a(ncc,ib1),q(ncc,ib3),bt(ntest,5),bp(npest,5),
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* spt(m,4),spp(m,4),h(ib3),wrk(lwrk)
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integer index(nrest),nummer(m)
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c ..local scalars..
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real*8 aa,acc,arg,cn,co,c1,dmax,d1,d2,eps,facc,facs,fac1,fac2,fn,
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* fpmax,fpms,f1,f2,f3,hti,htj,p,pi,pinv,piv,pi2,p1,p2,p3,ri,si,
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* sigma,sq,store,wi,rn,one,con1,con9,con4,half,ten
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integer i,iband,iband1,iband3,iband4,ich1,ich3,ii,ij,il,in,irot,
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* iter,i1,i2,i3,j,jlt,jrot,j1,j2,l,la,lf,lh,ll,lp,lt,lwest,l1,l2,
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* l3,l4,ncof,ncoff,npp,np4,nreg,nrint,nrr,nr1,ntt,nt4,nt6,num,
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* num1,rank
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c ..local arrays..
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real*8 ht(4),hp(4)
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c ..function references..
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real*8 abs,atan,fprati,sqrt,cos,sin
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integer min0
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c ..subroutine references..
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c fpback,fpbspl,fpgivs,fpdisc,fporde,fprank,fprota,fprpsp
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c ..
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c set constants
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one = 0.1e+01
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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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half = 0.5e0
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ten = 0.1e+02
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pi = atan(one)*4
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pi2 = pi+pi
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eps = sqrt(eta)
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if(iopt.lt.0) go to 70
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c calculation of acc, the absolute tolerance for the root of f(p)=s.
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acc = tol*s
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if(iopt.eq.0) go to 10
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if(s.lt.sup) then
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if (np.lt.11) go to 60
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go to 70
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endif
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c if iopt=0 we begin by computing the weighted least-squares polynomial
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c of the form
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c s(teta,phi) = c1*f1(teta) + cn*fn(teta)
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c where f1(teta) and fn(teta) are the cubic polynomials satisfying
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c f1(0) = 1, f1(pi) = f1'(0) = f1'(pi) = 0 ; fn(teta) = 1-f1(teta).
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c the corresponding weighted sum of squared residuals gives the upper
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c bound sup for the smoothing factor s.
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10 sup = 0.
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d1 = 0.
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d2 = 0.
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c1 = 0.
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cn = 0.
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fac1 = pi*(one + half)
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fac2 = (one + one)/pi**3
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aa = 0.
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do 40 i=1,m
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wi = w(i)
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ri = r(i)*wi
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arg = teta(i)
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fn = fac2*arg*arg*(fac1-arg)
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f1 = (one-fn)*wi
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fn = fn*wi
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if(fn.eq.0.) go to 20
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call fpgivs(fn,d1,co,si)
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call fprota(co,si,f1,aa)
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call fprota(co,si,ri,cn)
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20 if(f1.eq.0.) go to 30
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call fpgivs(f1,d2,co,si)
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call fprota(co,si,ri,c1)
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30 sup = sup+ri*ri
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40 continue
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if(d2.ne.0.) c1 = c1/d2
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if(d1.ne.0.) cn = (cn-aa*c1)/d1
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c find the b-spline representation of this least-squares polynomial
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nt = 8
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np = 8
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do 50 i=1,4
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c(i) = c1
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c(i+4) = c1
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c(i+8) = cn
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c(i+12) = cn
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tt(i) = 0.
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tt(i+4) = pi
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tp(i) = 0.
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tp(i+4) = pi2
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50 continue
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fp = sup
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c test whether the least-squares polynomial is an acceptable solution
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fpms = sup-s
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if(fpms.lt.acc) go to 960
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c test whether we cannot further increase the number of knots.
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60 if(npest.lt.11 .or. ntest.lt.9) go to 950
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c find the initial set of interior knots of the spherical spline in
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c case iopt = 0.
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np = 11
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tp(5) = pi*half
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tp(6) = pi
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tp(7) = tp(5)+pi
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nt = 9
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tt(5) = tp(5)
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cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
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c part 1 : computation of least-squares spherical splines. c
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c ******************************************************** c
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c if iopt < 0 we compute the least-squares spherical spline according c
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c to the given set of knots. c
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c if iopt >=0 we compute least-squares spherical splines with increas-c
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c ing numbers of knots until the corresponding sum f(p=inf)<=s. c
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c the initial set of knots then depends on the value of iopt: c
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c if iopt=0 we start with one interior knot in the teta-direction c
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c (pi/2) and three in the phi-direction (pi/2,pi,3*pi/2). c
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c if iopt>0 we start with the set of knots found at the last call c
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c of the routine. c
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cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
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c main loop for the different sets of knots. m is a save upper bound
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c for the number of trials.
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70 do 570 iter=1,m
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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(teta,phi).
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l1 = 4
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l2 = l1
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l3 = np-3
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l4 = l3
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tp(l2) = 0.
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tp(l3) = pi2
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do 80 i=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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tp(l2) = tp(l4)-pi2
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tp(l3) = tp(l1)+pi2
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80 continue
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l = nt
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do 90 i=1,4
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tt(i) = 0.
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tt(l) = pi
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l = l-1
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90 continue
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c find nrint, the total number of knot intervals and nreg, the number
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c of panels in which the approximation domain is subdivided by the
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c intersection of knots.
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ntt = nt-7
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npp = np-7
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nrr = npp/2
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nr1 = nrr+1
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nrint = ntt+npp
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nreg = ntt*npp
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c arrange the data points according to the panel they belong to.
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call fporde(teta,phi,m,3,3,tt,nt,tp,np,nummer,index,nreg)
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c find the b-spline coefficients coco and cosi of the cubic spline
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c approximations sc(phi) and ss(phi) for cos(phi) and sin(phi).
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do 100 i=1,npp
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coco(i) = 0.
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cosi(i) = 0.
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do 100 j=1,npp
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a(i,j) = 0.
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100 continue
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c the coefficients coco and cosi are obtained from the conditions
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c sc(tp(i))=cos(tp(i)),resp. ss(tp(i))=sin(tp(i)),i=4,5,...np-4.
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do 150 i=1,npp
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l2 = i+3
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arg = tp(l2)
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call fpbspl(tp,np,3,arg,l2,hp)
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do 110 j=1,npp
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row(j) = 0.
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110 continue
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ll = i
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do 120 j=1,3
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if(ll.gt.npp) ll= 1
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row(ll) = row(ll)+hp(j)
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ll = ll+1
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120 continue
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facc = cos(arg)
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facs = sin(arg)
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do 140 j=1,npp
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piv = row(j)
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if(piv.eq.0.) go to 140
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call fpgivs(piv,a(j,1),co,si)
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call fprota(co,si,facc,coco(j))
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call fprota(co,si,facs,cosi(j))
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if(j.eq.npp) go to 150
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j1 = j+1
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i2 = 1
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do 130 l=j1,npp
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i2 = i2+1
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call fprota(co,si,row(l),a(j,i2))
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130 continue
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140 continue
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150 continue
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call fpback(a,coco,npp,npp,coco,ncc)
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call fpback(a,cosi,npp,npp,cosi,ncc)
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c find ncof, the dimension of the spherical spline and ncoff, the
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c number of coefficients in the standard b-spline representation.
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nt4 = nt-4
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np4 = np-4
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ncoff = nt4*np4
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ncof = 6+npp*(ntt-1)
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c find the bandwidth of the observation matrix a.
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iband = 4*npp
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if(ntt.eq.4) iband = 3*(npp+1)
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if(ntt.lt.4) iband = ncof
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iband1 = iband-1
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c initialize the observation matrix a.
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do 160 i=1,ncof
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f(i) = 0.
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do 160 j=1,iband
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a(i,j) = 0.
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160 continue
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c initialize the sum of squared residuals.
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fp = 0.
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c fetch the data points in the new order. main loop for the
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c different panels.
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do 340 num=1,nreg
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c fix certain constants for the current panel; jrot records the column
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c number of the first non-zero element in a row of the observation
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c matrix according to a data point of the panel.
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num1 = num-1
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lt = num1/npp
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l1 = lt+4
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lp = num1-lt*npp+1
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l2 = lp+3
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lt = lt+1
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jrot = 0
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if(lt.gt.2) jrot = 3+(lt-3)*npp
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c test whether there are still data points in the current panel.
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in = index(num)
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170 if(in.eq.0) go to 340
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c fetch a new data point.
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wi = w(in)
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ri = r(in)*wi
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c evaluate for the teta-direction, the 4 non-zero b-splines at teta(in)
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call fpbspl(tt,nt,3,teta(in),l1,ht)
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c evaluate for the phi-direction, the 4 non-zero b-splines at phi(in)
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call fpbspl(tp,np,3,phi(in),l2,hp)
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c store the value of these b-splines in spt and spp resp.
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do 180 i=1,4
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spp(in,i) = hp(i)
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spt(in,i) = ht(i)
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180 continue
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c initialize the new row of observation matrix.
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do 190 i=1,iband
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h(i) = 0.
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190 continue
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c calculate the non-zero elements of the new row by making the cross
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c products of the non-zero b-splines in teta- and phi-direction and
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c by taking into account the conditions of the spherical splines.
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do 200 i=1,npp
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row(i) = 0.
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200 continue
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c take into account the condition (3) of the spherical splines.
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ll = lp
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do 210 i=1,4
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if(ll.gt.npp) ll=1
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row(ll) = row(ll)+hp(i)
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ll = ll+1
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210 continue
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c take into account the other conditions of the spherical splines.
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if(lt.gt.2 .and. lt.lt.(ntt-1)) go to 230
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facc = 0.
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facs = 0.
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do 220 i=1,npp
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facc = facc+row(i)*coco(i)
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facs = facs+row(i)*cosi(i)
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220 continue
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c fill in the non-zero elements of the new row.
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230 j1 = 0
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do 280 j =1,4
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jlt = j+lt
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htj = ht(j)
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if(jlt.gt.2 .and. jlt.le.nt4) go to 240
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j1 = j1+1
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h(j1) = h(j1)+htj
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go to 280
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240 if(jlt.eq.3 .or. jlt.eq.nt4) go to 260
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do 250 i=1,npp
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j1 = j1+1
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h(j1) = row(i)*htj
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250 continue
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go to 280
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260 if(jlt.eq.3) go to 270
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h(j1+1) = facc*htj
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h(j1+2) = facs*htj
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h(j1+3) = htj
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j1 = j1+2
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go to 280
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270 h(1) = h(1)+htj
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h(2) = facc*htj
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h(3) = facs*htj
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j1 = 3
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280 continue
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do 290 i=1,iband
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h(i) = h(i)*wi
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290 continue
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c rotate the row into triangle by givens transformations.
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irot = jrot
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do 310 i=1,iband
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irot = irot+1
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piv = h(i)
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if(piv.eq.0.) go to 310
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c calculate the parameters of the givens transformation.
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call fpgivs(piv,a(irot,1),co,si)
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c apply that transformation to the right hand side.
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call fprota(co,si,ri,f(irot))
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if(i.eq.iband) go to 320
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c apply that transformation to the left hand side.
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i2 = 1
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i3 = i+1
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do 300 j=i3,iband
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i2 = i2+1
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call fprota(co,si,h(j),a(irot,i2))
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300 continue
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310 continue
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c add the contribution of the row to the sum of squares of residual
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c right hand sides.
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320 fp = fp+ri**2
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c find the number of the next data point in the panel.
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in = nummer(in)
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go to 170
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340 continue
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c find dmax, the maximum value for the diagonal elements in the reduced
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c triangle.
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dmax = 0.
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do 350 i=1,ncof
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if(a(i,1).le.dmax) go to 350
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dmax = a(i,1)
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350 continue
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c check whether the observation matrix is rank deficient.
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sigma = eps*dmax
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do 360 i=1,ncof
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if(a(i,1).le.sigma) go to 370
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360 continue
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c backward substitution in case of full rank.
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call fpback(a,f,ncof,iband,c,ncc)
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rank = ncof
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do 365 i=1,ncof
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q(i,1) = a(i,1)/dmax
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365 continue
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go to 390
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c in case of rank deficiency, find the minimum norm solution.
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370 lwest = ncof*iband+ncof+iband
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if(lwrk.lt.lwest) go to 925
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lf = 1
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lh = lf+ncof
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la = lh+iband
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do 380 i=1,ncof
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ff(i) = f(i)
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do 380 j=1,iband
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q(i,j) = a(i,j)
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380 continue
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call fprank(q,ff,ncof,iband,ncc,sigma,c,sq,rank,wrk(la),
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* wrk(lf),wrk(lh))
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do 385 i=1,ncof
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q(i,1) = q(i,1)/dmax
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385 continue
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c add to the sum of squared residuals, the contribution of reducing
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c the rank.
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fp = fp+sq
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c find the coefficients in the standard b-spline representation of
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c the spherical spline.
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390 call fprpsp(nt,np,coco,cosi,c,ff,ncoff)
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c test whether the least-squares spline is an acceptable solution.
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if(iopt.lt.0) then
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if (fp.le.0) go to 970
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go to 980
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endif
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fpms = fp-s
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if(abs(fpms).le.acc) then
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if (fp.le.0) go to 970
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go to 980
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endif
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c if f(p=inf) < s, accept the choice of knots.
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if(fpms.lt.0.) go to 580
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c test whether we cannot further increase the number of knots.
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if(ncof.gt.m) go to 935
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c search where to add a new knot.
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c find for each interval the sum of squared residuals fpint for the
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c data points having the coordinate belonging to that knot interval.
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c calculate also coord which is the same sum, weighted by the position
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c of the data points considered.
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do 450 i=1,nrint
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fpint(i) = 0.
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coord(i) = 0.
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450 continue
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do 490 num=1,nreg
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num1 = num-1
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lt = num1/npp
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l1 = lt+1
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lp = num1-lt*npp
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l2 = lp+1+ntt
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jrot = lt*np4+lp
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in = index(num)
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460 if(in.eq.0) go to 490
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store = 0.
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i1 = jrot
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do 480 i=1,4
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hti = spt(in,i)
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j1 = i1
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do 470 j=1,4
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j1 = j1+1
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store = store+hti*spp(in,j)*c(j1)
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470 continue
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i1 = i1+np4
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480 continue
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store = (w(in)*(r(in)-store))**2
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fpint(l1) = fpint(l1)+store
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coord(l1) = coord(l1)+store*teta(in)
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fpint(l2) = fpint(l2)+store
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coord(l2) = coord(l2)+store*phi(in)
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in = nummer(in)
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go to 460
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490 continue
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c find the interval for which fpint is maximal on the condition that
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c there still can be added a knot.
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l1 = 1
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l2 = nrint
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if(ntest.lt.nt+1) l1=ntt+1
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if(npest.lt.np+2) l2=ntt
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c test whether we cannot further increase the number of knots.
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if(l1.gt.l2) go to 950
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500 fpmax = 0.
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l = 0
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do 510 i=l1,l2
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if(fpmax.ge.fpint(i)) go to 510
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l = i
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fpmax = fpint(i)
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510 continue
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if(l.eq.0) go to 930
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c calculate the position of the new knot.
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arg = coord(l)/fpint(l)
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c test in what direction the new knot is going to be added.
|
|
if(l.gt.ntt) go to 530
|
|
c addition in the teta-direction
|
|
l4 = l+4
|
|
fpint(l) = 0.
|
|
fac1 = tt(l4)-arg
|
|
fac2 = arg-tt(l4-1)
|
|
if(fac1.gt.(ten*fac2) .or. fac2.gt.(ten*fac1)) go to 500
|
|
j = nt
|
|
do 520 i=l4,nt
|
|
tt(j+1) = tt(j)
|
|
j = j-1
|
|
520 continue
|
|
tt(l4) = arg
|
|
nt = nt+1
|
|
go to 570
|
|
c addition in the phi-direction
|
|
530 l4 = l+4-ntt
|
|
if(arg.lt.pi) go to 540
|
|
arg = arg-pi
|
|
l4 = l4-nrr
|
|
540 fpint(l) = 0.
|
|
fac1 = tp(l4)-arg
|
|
fac2 = arg-tp(l4-1)
|
|
if(fac1.gt.(ten*fac2) .or. fac2.gt.(ten*fac1)) go to 500
|
|
ll = nrr+4
|
|
j = ll
|
|
do 550 i=l4,ll
|
|
tp(j+1) = tp(j)
|
|
j = j-1
|
|
550 continue
|
|
tp(l4) = arg
|
|
np = np+2
|
|
nrr = nrr+1
|
|
do 560 i=5,ll
|
|
j = i+nrr
|
|
tp(j) = tp(i)+pi
|
|
560 continue
|
|
c restart the computations with the new set of knots.
|
|
570 continue
|
|
cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
|
|
c part 2: determination of the smoothing spherical spline. c
|
|
c ******************************************************** c
|
|
c we have determined the number of knots and their position. we now c
|
|
c compute the coefficients of the smoothing spline sp(teta,phi). c
|
|
c the observation matrix a is extended by the rows of a matrix, expres-c
|
|
c sing that sp(teta,phi) must be a constant function in the variable c
|
|
c phi and a cubic polynomial in the variable teta. the corresponding c
|
|
c weights of these additional rows are set to 1/(p). iteratively c
|
|
c we than have to determine the value of p such that f(p) = sum((w(i)* c
|
|
c (r(i)-sp(teta(i),phi(i))))**2) be = s. c
|
|
c we already know that the least-squares polynomial corresponds to p=0,c
|
|
c and that the least-squares spherical spline corresponds to p=infin. c
|
|
c the iteration process makes use of rational interpolation. since f(p)c
|
|
c is a convex and strictly decreasing function of p, it can be approx- c
|
|
c imated by a rational function of the form r(p) = (u*p+v)/(p+w). c
|
|
c three values of p (p1,p2,p3) with corresponding values of f(p) (f1= c
|
|
c f(p1)-s,f2=f(p2)-s,f3=f(p3)-s) are used to calculate the new value c
|
|
c of p such that r(p)=s. convergence is guaranteed by taking f1>0,f3<0.c
|
|
cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
|
|
c evaluate the discontinuity jumps of the 3-th order derivative of
|
|
c the b-splines at the knots tt(l),l=5,...,nt-4.
|
|
580 call fpdisc(tt,nt,5,bt,ntest)
|
|
c evaluate the discontinuity jumps of the 3-th order derivative of
|
|
c the b-splines at the knots tp(l),l=5,...,np-4.
|
|
call fpdisc(tp,np,5,bp,npest)
|
|
c initial value for p.
|
|
p1 = 0.
|
|
f1 = sup-s
|
|
p3 = -one
|
|
f3 = fpms
|
|
p = 0.
|
|
do 585 i=1,ncof
|
|
p = p+a(i,1)
|
|
585 continue
|
|
rn = ncof
|
|
p = rn/p
|
|
c find the bandwidth of the extended observation matrix.
|
|
iband4 = iband+3
|
|
if(ntt.le.4) iband4 = ncof
|
|
iband3 = iband4 -1
|
|
ich1 = 0
|
|
ich3 = 0
|
|
c iteration process to find the root of f(p)=s.
|
|
do 920 iter=1,maxit
|
|
pinv = one/p
|
|
c store the triangularized observation matrix into q.
|
|
do 600 i=1,ncof
|
|
ff(i) = f(i)
|
|
do 590 j=1,iband4
|
|
q(i,j) = 0.
|
|
590 continue
|
|
do 600 j=1,iband
|
|
q(i,j) = a(i,j)
|
|
600 continue
|
|
c extend the observation matrix with the rows of a matrix, expressing
|
|
c that for teta=cst. sp(teta,phi) must be a constant function.
|
|
nt6 = nt-6
|
|
do 720 i=5,np4
|
|
ii = i-4
|
|
do 610 l=1,npp
|
|
row(l) = 0.
|
|
610 continue
|
|
ll = ii
|
|
do 620 l=1,5
|
|
if(ll.gt.npp) ll=1
|
|
row(ll) = row(ll)+bp(ii,l)
|
|
ll = ll+1
|
|
620 continue
|
|
facc = 0.
|
|
facs = 0.
|
|
do 630 l=1,npp
|
|
facc = facc+row(l)*coco(l)
|
|
facs = facs+row(l)*cosi(l)
|
|
630 continue
|
|
do 720 j=1,nt6
|
|
c initialize the new row.
|
|
do 640 l=1,iband
|
|
h(l) = 0.
|
|
640 continue
|
|
c fill in the non-zero elements of the row. jrot records the column
|
|
c number of the first non-zero element in the row.
|
|
jrot = 4+(j-2)*npp
|
|
if(j.gt.1 .and. j.lt.nt6) go to 650
|
|
h(1) = facc
|
|
h(2) = facs
|
|
if(j.eq.1) jrot = 2
|
|
go to 670
|
|
650 do 660 l=1,npp
|
|
h(l)=row(l)
|
|
660 continue
|
|
670 do 675 l=1,iband
|
|
h(l) = h(l)*pinv
|
|
675 continue
|
|
ri = 0.
|
|
c rotate the new row into triangle by givens transformations.
|
|
do 710 irot=jrot,ncof
|
|
piv = h(1)
|
|
i2 = min0(iband1,ncof-irot)
|
|
if(piv.eq.0.) then
|
|
if (i2.le.0) go to 720
|
|
go to 690
|
|
endif
|
|
c calculate the parameters of the givens transformation.
|
|
call fpgivs(piv,q(irot,1),co,si)
|
|
c apply that givens transformation to the right hand side.
|
|
call fprota(co,si,ri,ff(irot))
|
|
if(i2.eq.0) go to 720
|
|
c apply that givens transformation to the left hand side.
|
|
do 680 l=1,i2
|
|
l1 = l+1
|
|
call fprota(co,si,h(l1),q(irot,l1))
|
|
680 continue
|
|
690 do 700 l=1,i2
|
|
h(l) = h(l+1)
|
|
700 continue
|
|
h(i2+1) = 0.
|
|
710 continue
|
|
720 continue
|
|
c extend the observation matrix with the rows of a matrix expressing
|
|
c that for phi=cst. sp(teta,phi) must be a cubic polynomial.
|
|
do 810 i=5,nt4
|
|
ii = i-4
|
|
do 810 j=1,npp
|
|
c initialize the new row
|
|
do 730 l=1,iband4
|
|
h(l) = 0.
|
|
730 continue
|
|
c fill in the non-zero elements of the row. jrot records the column
|
|
c number of the first non-zero element in the row.
|
|
j1 = 1
|
|
do 760 l=1,5
|
|
il = ii+l
|
|
ij = npp
|
|
if(il.ne.3 .and. il.ne.nt4) go to 750
|
|
j1 = j1+3-j
|
|
j2 = j1-2
|
|
ij = 0
|
|
if(il.ne.3) go to 740
|
|
j1 = 1
|
|
j2 = 2
|
|
ij = j+2
|
|
740 h(j2) = bt(ii,l)*coco(j)
|
|
h(j2+1) = bt(ii,l)*cosi(j)
|
|
750 h(j1) = h(j1)+bt(ii,l)
|
|
j1 = j1+ij
|
|
760 continue
|
|
do 765 l=1,iband4
|
|
h(l) = h(l)*pinv
|
|
765 continue
|
|
ri = 0.
|
|
jrot = 1
|
|
if(ii.gt.2) jrot = 3+j+(ii-3)*npp
|
|
c rotate the new row into triangle by givens transformations.
|
|
do 800 irot=jrot,ncof
|
|
piv = h(1)
|
|
i2 = min0(iband3,ncof-irot)
|
|
if(piv.eq.0.) then
|
|
if (i2.le.0) go to 810
|
|
go to 780
|
|
endif
|
|
c calculate the parameters of the givens transformation.
|
|
call fpgivs(piv,q(irot,1),co,si)
|
|
c apply that givens transformation to the right hand side.
|
|
call fprota(co,si,ri,ff(irot))
|
|
if(i2.eq.0) go to 810
|
|
c apply that givens transformation to the left hand side.
|
|
do 770 l=1,i2
|
|
l1 = l+1
|
|
call fprota(co,si,h(l1),q(irot,l1))
|
|
770 continue
|
|
780 do 790 l=1,i2
|
|
h(l) = h(l+1)
|
|
790 continue
|
|
h(i2+1) = 0.
|
|
800 continue
|
|
810 continue
|
|
c find dmax, the maximum value for the diagonal elements in the
|
|
c reduced triangle.
|
|
dmax = 0.
|
|
do 820 i=1,ncof
|
|
if(q(i,1).le.dmax) go to 820
|
|
dmax = q(i,1)
|
|
820 continue
|
|
c check whether the matrix is rank deficient.
|
|
sigma = eps*dmax
|
|
do 830 i=1,ncof
|
|
if(q(i,1).le.sigma) go to 840
|
|
830 continue
|
|
c backward substitution in case of full rank.
|
|
call fpback(q,ff,ncof,iband4,c,ncc)
|
|
rank = ncof
|
|
go to 845
|
|
c in case of rank deficiency, find the minimum norm solution.
|
|
840 lwest = ncof*iband4+ncof+iband4
|
|
if(lwrk.lt.lwest) go to 925
|
|
lf = 1
|
|
lh = lf+ncof
|
|
la = lh+iband4
|
|
call fprank(q,ff,ncof,iband4,ncc,sigma,c,sq,rank,wrk(la),
|
|
* wrk(lf),wrk(lh))
|
|
845 do 850 i=1,ncof
|
|
q(i,1) = q(i,1)/dmax
|
|
850 continue
|
|
c find the coefficients in the standard b-spline representation of
|
|
c the spherical spline.
|
|
call fprpsp(nt,np,coco,cosi,c,ff,ncoff)
|
|
c compute f(p).
|
|
fp = 0.
|
|
do 890 num = 1,nreg
|
|
num1 = num-1
|
|
lt = num1/npp
|
|
lp = num1-lt*npp
|
|
jrot = lt*np4+lp
|
|
in = index(num)
|
|
860 if(in.eq.0) go to 890
|
|
store = 0.
|
|
i1 = jrot
|
|
do 880 i=1,4
|
|
hti = spt(in,i)
|
|
j1 = i1
|
|
do 870 j=1,4
|
|
j1 = j1+1
|
|
store = store+hti*spp(in,j)*c(j1)
|
|
870 continue
|
|
i1 = i1+np4
|
|
880 continue
|
|
fp = fp+(w(in)*(r(in)-store))**2
|
|
in = nummer(in)
|
|
go to 860
|
|
890 continue
|
|
c test whether the approximation sp(teta,phi) is an acceptable solution
|
|
fpms = fp-s
|
|
if(abs(fpms).le.acc) go to 980
|
|
c test whether the maximum allowable number of iterations has been
|
|
c reached.
|
|
if(iter.eq.maxit) go to 940
|
|
c carry out one more step of the iteration process.
|
|
p2 = p
|
|
f2 = fpms
|
|
if(ich3.ne.0) go to 900
|
|
if((f2-f3).gt.acc) go to 895
|
|
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 920
|
|
895 if(f2.lt.0.) ich3 = 1
|
|
900 if(ich1.ne.0) go to 910
|
|
if((f1-f2).gt.acc) go to 905
|
|
c our initial choice of p is too small
|
|
p1 = p2
|
|
f1 = f2
|
|
p = p/con4
|
|
if(p3.lt.0.) go to 920
|
|
if(p.ge.p3) p = p2*con1 +p3*con9
|
|
go to 920
|
|
905 if(f2.gt.0.) ich1 = 1
|
|
c test whether the iteration process proceeds as theoretically
|
|
c expected.
|
|
910 if(f2.ge.f1 .or. f2.le.f3) go to 945
|
|
c find the new value of p.
|
|
p = fprati(p1,f1,p2,f2,p3,f3)
|
|
920 continue
|
|
c error codes and messages.
|
|
925 ier = lwest
|
|
go to 990
|
|
930 ier = 5
|
|
go to 990
|
|
935 ier = 4
|
|
go to 990
|
|
940 ier = 3
|
|
go to 990
|
|
945 ier = 2
|
|
go to 990
|
|
950 ier = 1
|
|
go to 990
|
|
960 ier = -2
|
|
go to 990
|
|
970 ier = -1
|
|
fp = 0.
|
|
980 if(ncof.ne.rank) ier = -rank
|
|
990 return
|
|
end
|