subroutine fpcurf(iopt,x,y,w,m,xb,xe,k,s,nest,tol,maxit,k1,k2,
* n,t,c,fp,fpint,z,a,b,g,q,nrdata,ier)
c ..
c ..scalar arguments..
real xb,xe,s,tol,fp
integer iopt,m,k,nest,maxit,k1,k2,n,ier
c ..array arguments..
real x(m),y(m),w(m),t(nest),c(nest),fpint(nest),
* z(nest),a(nest,k1),b(nest,k2),g(nest,k2),q(m,k1)
integer nrdata(nest)
c ..local scalars..
real acc,con1,con4,con9,cos,half,fpart,fpms,fpold,fp0,f1,f2,f3,
* one,p,pinv,piv,p1,p2,p3,rn,sin,store,term,wi,xi,yi
integer i,ich1,ich3,it,iter,i1,i2,i3,j,k3,l,l0,
* mk1,new,nk1,nmax,nmin,nplus,npl1,nrint,n8
c ..local arrays..
real h(7)
c ..function references
real abs,fprati
integer max0,min0
c ..subroutine references..
c fpback,fpbspl,fpgivs,fpdisc,fpknot,fprota
c ..
c set constants
one = 0.1e+01
con1 = 0.1e0
con9 = 0.9e0
con4 = 0.4e-01
half = 0.5e0
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(x), c
c and the corresponding sum of squared residuals fp=f(p=inf). c
c if iopt=-1 sinf(x) 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 nmax = m+k+1. c
c if s > 0 and c
c iopt=0 we first compute the least-squares polynomial of c
c degree k; n = nmin = 2*k+2 c
c iopt=1 we start with the set of knots found at the last c
c call of the routine, except for the case that s > fp0; then c
c we compute directly the least-squares polynomial of degree k. c
cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
c determine nmin, the number of knots for polynomial approximation.
nmin = 2*k1
if(iopt.lt.0) go to 60
c calculation of acc, the absolute tolerance for the root of f(p)=s.
acc = tol*s
c determine nmax, the number of knots for spline interpolation.
nmax = m+k1
if(s.gt.0.) go to 45
c if s=0, s(x) is an interpolating spline.
c test whether the required storage space exceeds the available one.
n = nmax
if(nmax.gt.nest) go to 420
c find the position of the interior knots in case of interpolation.
10 mk1 = m-k1
if(mk1.eq.0) go to 60
k3 = k/2
i = k2
j = k3+2
if(k3*2.eq.k) go to 30
do 20 l=1,mk1
t(i) = x(j)
i = i+1
j = j+1
20 continue
go to 60
30 do 40 l=1,mk1
t(i) = (x(j)+x(j-1))*half
i = i+1
j = j+1
40 continue
go to 60
c if s>0 our initial choice of knots depends on the value of iopt.
c if iopt=0 or iopt=1 and s>=fp0, we start computing the least-squares
c polynomial of degree k which is a spline without interior knots.
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.
45 if(iopt.eq.0) go to 50
if(n.eq.nmin) go to 50
fp0 = fpint(n)
fpold = fpint(n-1)
nplus = nrdata(n)
if(fp0.gt.s) go to 60
50 n = nmin
fpold = 0.
nplus = 0
nrdata(1) = m-2
c main loop for the different sets of knots. m is a save upper bound
c for the number of trials.
60 do 200 iter = 1,m
if(n.eq.nmin) ier = -2
c find nrint, tne number of knot intervals.
nrint = n-nmin+1
c find the position of the additional knots which are needed for
c the b-spline representation of s(x).
nk1 = n-k1
i = n
do 70 j=1,k1
t(j) = xb
t(i) = xe
i = i-1
70 continue
c compute the b-spline coefficients of the least-squares spline
c sinf(x). the observation matrix a is built up row by row and
c reduced to upper triangular form by givens transformations.
c at the same time fp=f(p=inf) is computed.
fp = 0.
c initialize the observation matrix a.
do 80 i=1,nk1
z(i) = 0.
do 80 j=1,k1
a(i,j) = 0.
80 continue
l = k1
do 130 it=1,m
c fetch the current data point x(it),y(it).
xi = x(it)
wi = w(it)
yi = y(it)*wi
c search for knot interval t(l) <= xi < t(l+1).
85 if(xi.lt.t(l+1) .or. l.eq.nk1) go to 90
l = l+1
go to 85
c evaluate the (k+1) non-zero b-splines at xi and store them in q.
90 call fpbspl(t,n,k,xi,l,h)
do 95 i=1,k1
q(it,i) = h(i)
h(i) = h(i)*wi
95 continue
c rotate the new row of the observation matrix into triangle.
j = l-k1
do 110 i=1,k1
j = j+1
piv = h(i)
if(piv.eq.0.) go to 110
c calculate the parameters of the givens transformation.
call fpgivs(piv,a(j,1),cos,sin)
c transformations to right hand side.
call fprota(cos,sin,yi,z(j))
if(i.eq.k1) go to 120
i2 = 1
i3 = i+1
do 100 i1 = i3,k1
i2 = i2+1
c transformations to left hand side.
call fprota(cos,sin,h(i1),a(j,i2))
100 continue
110 continue
c add contribution of this row to the sum of squares of residual
c right hand sides.
120 fp = fp+yi**2
130 continue
if(ier.eq.(-2)) fp0 = fp
fpint(n) = fp0
fpint(n-1) = fpold
nrdata(n) = nplus
c backward substitution to obtain the b-spline coefficients.
call fpback(a,z,nk1,k1,c,nest)
c test whether the approximation sinf(x) 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 accept the choice of knots.
if(fpms.lt.0.) go to 250
c if n = nmax, sinf(x) is an interpolating spline.
if(n.eq.nmax) go to 430
c increase the number of knots.
c if n=nest we cannot increase the number of knots because of
c the storage capacity limitation.
if(n.eq.nest) go to 420
c determine the number of knots nplus we are going to add.
if(ier.eq.0) go to 140
nplus = 1
ier = 0
go to 150
140 npl1 = nplus*2
rn = nplus
if(fpold-fp.gt.acc) npl1 = rn*fpms/(fpold-fp)
nplus = min0(nplus*2,max0(npl1,nplus/2,1))
150 fpold = fp
c compute the sum((w(i)*(y(i)-s(x(i))))**2) for each knot interval
c t(j+k) <= x(i) <= t(j+k+1) and store it in fpint(j),j=1,2,...nrint.
fpart = 0.
i = 1
l = k2
new = 0
do 180 it=1,m
if(x(it).lt.t(l) .or. l.gt.nk1) go to 160
new = 1
l = l+1
160 term = 0.
l0 = l-k2
do 170 j=1,k1
l0 = l0+1
term = term+c(l0)*q(it,j)
170 continue
term = (w(it)*(term-y(it)))**2
fpart = fpart+term
if(new.eq.0) go to 180
store = term*half
fpint(i) = fpart-store
i = i+1
fpart = store
new = 0
180 continue
fpint(nrint) = fpart
do 190 l=1,nplus
c add a new knot.
call fpknot(x,m,t,n,fpint,nrdata,nrint,nest,1)
c if n=nmax we locate the knots as for interpolation.
if(n.eq.nmax) go to 10
c test whether we cannot further increase the number of knots.
if(n.eq.nest) go to 200
190 continue
c restart the computations with the new set of knots.
200 continue
c test whether the least-squares kth degree polynomial is a solution
c of our approximation problem.
250 if(ier.eq.(-2)) go to 440
cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
c part 2: determination of the smoothing spline sp(x). c
c *************************************************** c
c we have determined the number of knots and their position. c
c we now compute the b-spline coefficients of the smoothing spline c
c sp(x). the observation matrix a is extended by the rows of matrix c
c b expressing that the kth derivative discontinuities of sp(x) at c
c the interior knots t(k+2),...t(n-k-1) must be zero. the corres- c
c ponding weights of these additional rows are set to 1/p. c
c iteratively we then have to determine the value of p such that c
c f(p)=sum((w(i)*(y(i)-sp(x(i))))**2) be = s. we already know that c
c the least-squares kth degree polynomial corresponds to p=0, and c
c that the least-squares spline corresponds to p=infinity. the c
c iteration process which is proposed here, makes use of rational c
c interpolation. since f(p) is a convex and strictly decreasing c
c function of p, it can be approximated by a rational function c
c r(p) = (u*p+v)/(p+w). three values of p(p1,p2,p3) with correspond- c
c ing values of f(p) (f1=f(p1)-s,f2=f(p2)-s,f3=f(p3)-s) are used c
c to calculate the new value of p such that r(p)=s. convergence is c
c guaranteed by taking f1>0 and f3<0. c
cccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc
c evaluate the discontinuity jump of the kth derivative of the
c b-splines at the knots t(l),l=k+2,...n-k-1 and store in b.
call fpdisc(t,n,k2,b,nest)
c initial value for p.
p1 = 0.
f1 = fp0-s
p3 = -one
f3 = fpms
p = 0.
do 255 i=1,nk1
p = p+a(i,1)
255 continue
rn = nk1
p = rn/p
ich1 = 0
ich3 = 0
n8 = n-nmin
c iteration process to find the root of f(p) = s.
do 360 iter=1,maxit
c the rows of matrix b with weight 1/p are rotated into the
c triangularised observation matrix a which is stored in g.
pinv = one/p
do 260 i=1,nk1
c(i) = z(i)
g(i,k2) = 0.
do 260 j=1,k1
g(i,j) = a(i,j)
260 continue
do 300 it=1,n8
c the row of matrix b is rotated into triangle by givens transformation
do 270 i=1,k2
h(i) = b(it,i)*pinv
270 continue
yi = 0.
do 290 j=it,nk1
piv = h(1)
c calculate the parameters of the givens transformation.
call fpgivs(piv,g(j,1),cos,sin)
c transformations to right hand side.
call fprota(cos,sin,yi,c(j))
if(j.eq.nk1) go to 300
i2 = k1
if(j.gt.n8) i2 = nk1-j
do 280 i=1,i2
c transformations to left hand side.
i1 = i+1
call fprota(cos,sin,h(i1),g(j,i1))
h(i) = h(i1)
280 continue
h(i2+1) = 0.
290 continue
300 continue
c backward substitution to obtain the b-spline coefficients.
call fpback(g,c,nk1,k2,c,nest)
c computation of f(p).
fp = 0.
l = k2
do 330 it=1,m
if(x(it).lt.t(l) .or. l.gt.nk1) go to 310
l = l+1
310 l0 = l-k2
term = 0.
do 320 j=1,k1
l0 = l0+1
term = term+c(l0)*q(it,j)
320 continue
fp = fp+(w(it)*(term-y(it)))**2
330 continue
c test whether the approximation sp(x) is an acceptable solution.
fpms = fp-s
if(abs(fpms).lt.acc) go to 440
c test whether the maximal number of iterations is 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 340
if((f2-f3).gt.acc) go to 335
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 360
335 if(f2.lt.0.) ich3=1
340 if(ich1.ne.0) go to 350
if((f1-f2).gt.acc) go to 345
c our initial choice of p is too small
p1 = p2
f1 = f2
p = p/con4
if(p3.lt.0.) go to 360
if(p.ge.p3) p = p2*con1 + p3*con9
go to 360
345 if(f2.gt.0.) ich1=1
c test whether the iteration process proceeds as theoretically
c expected.
350 if(f2.ge.f1 .or. f2.le.f3) go to 410
c find the new value for p.
p = fprati(p1,f1,p2,f2,p3,f3)
360 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
440 return
end