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mountcontrol/cxx/fitpack/fpgrdi.f
Timur A. Fatkhullin 5279d1c41a add FITPACK Fortran library 
start developing of FITPACK C++ bindings
mount_server.cpp: fix compilation error with GCC15
2025-05-05 17:24:21 +03:00

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recursive subroutine fpgrdi(ifsu,ifsv,ifbu,ifbv,iback,u,mu,v,
* mv,z,mz,dz,iop0,iop1,tu,nu,tv,nv,p,c,nc,sq,fp,fpu,fpv,mm,
* mvnu,spu,spv,right,q,au,av1,av2,bu,bv,aa,bb,cc,cosi,nru,nrv)
implicit none
c ..
c ..scalar arguments..
real*8 p,sq,fp
integer ifsu,ifsv,ifbu,ifbv,iback,mu,mv,mz,iop0,iop1,nu,nv,nc,
* mm,mvnu
c ..array arguments..
real*8 u(mu),v(mv),z(mz),dz(3),tu(nu),tv(nv),c(nc),fpu(nu),fpv(nv)
*,
* spu(mu,4),spv(mv,4),right(mm),q(mvnu),au(nu,5),av1(nv,6),
* av2(nv,4),aa(2,mv),bb(2,nv),cc(nv),cosi(2,nv),bu(nu,5),bv(nv,5)
integer nru(mu),nrv(mv)
c ..local scalars..
real*8 arg,co,dz1,dz2,dz3,fac,fac0,pinv,piv,si,term,one,three,half
*
integer i,ic,ii,ij,ik,iq,irot,it,iz,i0,i1,i2,i3,j,jj,jk,jper,
* j0,j1,k,k1,k2,l,l0,l1,l2,mvv,ncof,nrold,nroldu,nroldv,number,
* numu,numu1,numv,numv1,nuu,nu4,nu7,nu8,nu9,nv11,nv4,nv7,nv8,n1
c ..local arrays..
real*8 h(5),h1(5),h2(4)
c ..function references..
integer min0
real*8 cos,sin
c ..subroutine references..
c fpback,fpbspl,fpgivs,fpcyt1,fpcyt2,fpdisc,fpbacp,fprota
c ..
c let
c | (spu) | | (spv) |
c (au) = | ---------- | (av) = | ---------- |
c | (1/p) (bu) | | (1/p) (bv) |
c
c | z ' 0 |
c q = | ------ |
c | 0 ' 0 |
c
c with c : the (nu-4) x (nv-4) matrix which contains the b-spline
c coefficients.
c z : the mu x mv matrix which contains the function values.
c spu,spv: the mu x (nu-4), resp. mv x (nv-4) observation matrices
c according to the least-squares problems in the u-,resp.
c v-direction.
c bu,bv : the (nu-7) x (nu-4),resp. (nv-7) x (nv-4) matrices
c containing the discontinuity jumps of the derivatives
c of the b-splines in the u-,resp.v-variable at the knots
c the b-spline coefficients of the smoothing spline are then calculated
c as the least-squares solution of the following over-determined linear
c system of equations
c
c (1) (av) c (au)' = q
c
c subject to the constraints
c
c (2) c(i,nv-3+j) = c(i,j), j=1,2,3 ; i=1,2,...,nu-4
c
c (3) if iop0 = 0 c(1,j) = dz(1)
c iop0 = 1 c(1,j) = dz(1)
c c(2,j) = dz(1)+(dz(2)*cosi(1,j)+dz(3)*cosi(2,j))*
c tu(5)/3. = cc(j) , j=1,2,...nv-4
c
c (4) if iop1 = 1 c(nu-4,j) = 0, j=1,2,...,nv-4.
c
c set constants
one = 1
three = 3
half = 0.5
c initialization
nu4 = nu-4
nu7 = nu-7
nu8 = nu-8
nu9 = nu-9
nv4 = nv-4
nv7 = nv-7
nv8 = nv-8
nv11 = nv-11
nuu = nu4-iop0-iop1-1
if(p.gt.0.) pinv = one/p
c it depends on the value of the flags ifsu,ifsv,ifbu,ifbv and iop0 and
c on the value of p whether the matrices (spu), (spv), (bu), (bv) and
c (cosi) still must be determined.
if(ifsu.ne.0) go to 30
c calculate the non-zero elements of the matrix (spu) which is the ob-
c servation matrix according to the least-squares spline approximation
c problem in the u-direction.
l = 4
l1 = 5
number = 0
do 25 it=1,mu
arg = u(it)
10 if(arg.lt.tu(l1) .or. l.eq.nu4) go to 15
l = l1
l1 = l+1
number = number+1
go to 10
15 call fpbspl(tu,nu,3,arg,l,h)
do 20 i=1,4
spu(it,i) = h(i)
20 continue
nru(it) = number
25 continue
ifsu = 1
c calculate the non-zero elements of the matrix (spv) which is the ob-
c servation matrix according to the least-squares spline approximation
c problem in the v-direction.
30 if(ifsv.ne.0) go to 85
l = 4
l1 = 5
number = 0
do 50 it=1,mv
arg = v(it)
35 if(arg.lt.tv(l1) .or. l.eq.nv4) go to 40
l = l1
l1 = l+1
number = number+1
go to 35
40 call fpbspl(tv,nv,3,arg,l,h)
do 45 i=1,4
spv(it,i) = h(i)
45 continue
nrv(it) = number
50 continue
ifsv = 1
if(iop0.eq.0) go to 85
c calculate the coefficients of the interpolating splines for cos(v)
c and sin(v).
do 55 i=1,nv4
cosi(1,i) = 0.
cosi(2,i) = 0.
55 continue
if(nv7.lt.4) go to 85
do 65 i=1,nv7
l = i+3
arg = tv(l)
call fpbspl(tv,nv,3,arg,l,h)
do 60 j=1,3
av1(i,j) = h(j)
60 continue
cosi(1,i) = cos(arg)
cosi(2,i) = sin(arg)
65 continue
call fpcyt1(av1,nv7,nv)
do 80 j=1,2
do 70 i=1,nv7
right(i) = cosi(j,i)
70 continue
call fpcyt2(av1,nv7,right,right,nv)
do 75 i=1,nv7
cosi(j,i+1) = right(i)
75 continue
cosi(j,1) = cosi(j,nv7+1)
cosi(j,nv7+2) = cosi(j,2)
cosi(j,nv4) = cosi(j,3)
80 continue
85 if(p.le.0.) go to 150
c calculate the non-zero elements of the matrix (bu).
if(ifbu.ne.0 .or. nu8.eq.0) go to 90
call fpdisc(tu,nu,5,bu,nu)
ifbu = 1
c calculate the non-zero elements of the matrix (bv).
90 if(ifbv.ne.0 .or. nv8.eq.0) go to 150
call fpdisc(tv,nv,5,bv,nv)
ifbv = 1
c substituting (2),(3) and (4) into (1), we obtain the overdetermined
c system
c (5) (avv) (cr) (auu)' = (qq)
c from which the nuu*nv7 remaining coefficients
c c(i,j) , i=2+iop0,3+iop0,...,nu-4-iop1 ; j=1,2,...,nv-7 ,
c the elements of (cr), are then determined in the least-squares sense.
c simultaneously, we compute the resulting sum of squared residuals sq.
150 dz1 = dz(1)
do 155 i=1,mv
aa(1,i) = dz1
155 continue
if(nv8.eq.0 .or. p.le.0.) go to 165
do 160 i=1,nv8
bb(1,i) = 0.
160 continue
165 mvv = mv
if(iop0.eq.0) go to 220
fac = tu(5)/three
dz2 = dz(2)*fac
dz3 = dz(3)*fac
do 170 i=1,nv4
cc(i) = dz1+dz2*cosi(1,i)+dz3*cosi(2,i)
170 continue
do 190 i=1,mv
number = nrv(i)
fac = 0.
do 180 j=1,4
number = number+1
fac = fac+cc(number)*spv(i,j)
180 continue
aa(2,i) = fac
190 continue
if(nv8.eq.0 .or. p.le.0.) go to 220
do 210 i=1,nv8
number = i
fac = 0.
do 200 j=1,5
fac = fac+cc(number)*bv(i,j)
number = number+1
200 continue
bb(2,i) = fac*pinv
210 continue
mvv = mvv+nv8
c we first determine the matrices (auu) and (qq). then we reduce the
c matrix (auu) to upper triangular form (ru) using givens rotations.
c we apply the same transformations to the rows of matrix qq to obtain
c the (mv+nv8) x nuu matrix g.
c we store matrix (ru) into au and g into q.
220 l = mvv*nuu
c initialization.
sq = 0.
do 230 i=1,l
q(i) = 0.
230 continue
do 240 i=1,nuu
do 240 j=1,5
au(i,j) = 0.
240 continue
l = 0
nrold = 0
n1 = nrold+1
do 420 it=1,mu
number = nru(it)
c find the appropriate column of q.
250 do 260 j=1,mvv
right(j) = 0.
260 continue
if(nrold.eq.number) go to 280
if(p.le.0.) go to 410
c fetch a new row of matrix (bu).
do 270 j=1,5
h(j) = bu(n1,j)*pinv
270 continue
i0 = 1
i1 = 5
go to 310
c fetch a new row of matrix (spu).
280 do 290 j=1,4
h(j) = spu(it,j)
290 continue
c find the appropriate column of q.
do 300 j=1,mv
l = l+1
right(j) = z(l)
300 continue
i0 = 1
i1 = 4
310 if(nu7-number .eq. iop1) i1 = i1-1
j0 = n1
c take into account that we eliminate the constraints (3)
320 if(j0-1.gt.iop0) go to 360
fac0 = h(i0)
do 330 j=1,mv
right(j) = right(j)-fac0*aa(j0,j)
330 continue
if(mv.eq.mvv) go to 350
j = mv
do 340 jj=1,nv8
j = j+1
right(j) = right(j)-fac0*bb(j0,jj)
340 continue
350 j0 = j0+1
i0 = i0+1
go to 320
360 irot = nrold-iop0-1
if(irot.lt.0) irot = 0
c rotate the new row of matrix (auu) into triangle.
do 390 i=i0,i1
irot = irot+1
piv = h(i)
if(piv.eq.0.) go to 390
c calculate the parameters of the givens transformation.
call fpgivs(piv,au(irot,1),co,si)
c apply that transformation to the rows of matrix (qq).
iq = (irot-1)*mvv
do 370 j=1,mvv
iq = iq+1
call fprota(co,si,right(j),q(iq))
370 continue
c apply that transformation to the columns of (auu).
if(i.eq.i1) go to 390
i2 = 1
i3 = i+1
do 380 j=i3,i1
i2 = i2+1
call fprota(co,si,h(j),au(irot,i2))
380 continue
390 continue
c we update the sum of squared residuals
do 395 j=1,mvv
sq = sq+right(j)**2
395 continue
if(nrold.eq.number) go to 420
410 nrold = n1
n1 = n1+1
go to 250
420 continue
c we determine the matrix (avv) and then we reduce her to
c upper triangular form (rv) using givens rotations.
c we apply the same transformations to the columns of matrix
c g to obtain the (nv-7) x (nu-5-iop0-iop1) matrix h.
c we store matrix (rv) into av1 and av2, h into c.
c the nv7 x nv7 upper triangular matrix (rv) has the form
c | av1 ' |
c (rv) = | ' av2 |
c | 0 ' |
c with (av2) a nv7 x 4 matrix and (av1) a nv11 x nv11 upper
c triangular matrix of bandwidth 5.
ncof = nuu*nv7
c initialization.
do 430 i=1,ncof
c(i) = 0.
430 continue
do 440 i=1,nv4
av1(i,5) = 0.
do 440 j=1,4
av1(i,j) = 0.
av2(i,j) = 0.
440 continue
jper = 0
nrold = 0
do 770 it=1,mv
number = nrv(it)
450 if(nrold.eq.number) go to 480
if(p.le.0.) go to 760
c fetch a new row of matrix (bv).
n1 = nrold+1
do 460 j=1,5
h(j) = bv(n1,j)*pinv
460 continue
c find the appropriate row of g.
do 465 j=1,nuu
right(j) = 0.
465 continue
if(mv.eq.mvv) go to 510
l = mv+n1
do 470 j=1,nuu
right(j) = q(l)
l = l+mvv
470 continue
go to 510
c fetch a new row of matrix (spv)
480 h(5) = 0.
do 490 j=1,4
h(j) = spv(it,j)
490 continue
c find the appropriate row of g.
l = it
do 500 j=1,nuu
right(j) = q(l)
l = l+mvv
500 continue
c test whether there are non-zero values in the new row of (avv)
c corresponding to the b-splines n(j,v),j=nv7+1,...,nv4.
510 if(nrold.lt.nv11) go to 710
if(jper.ne.0) go to 550
c initialize the matrix (av2).
jk = nv11+1
do 540 i=1,4
ik = jk
do 520 j=1,5
if(ik.le.0) go to 530
av2(ik,i) = av1(ik,j)
ik = ik-1
520 continue
530 jk = jk+1
540 continue
jper = 1
c if one of the non-zero elements of the new row corresponds to one of
c the b-splines n(j;v),j=nv7+1,...,nv4, we take account of condition
c (2) for setting up this row of (avv). the row is stored in h1( the
c part with respect to av1) and h2 (the part with respect to av2).
550 do 560 i=1,4
h1(i) = 0.
h2(i) = 0.
560 continue
h1(5) = 0.
j = nrold-nv11
do 600 i=1,5
j = j+1
l0 = j
570 l1 = l0-4
if(l1.le.0) go to 590
if(l1.le.nv11) go to 580
l0 = l1-nv11
go to 570
580 h1(l1) = h(i)
go to 600
590 h2(l0) = h2(l0) + h(i)
600 continue
c rotate the new row of (avv) into triangle.
if(nv11.le.0) go to 670
c rotations with the rows 1,2,...,nv11 of (avv).
do 660 j=1,nv11
piv = h1(1)
i2 = min0(nv11-j,4)
if(piv.eq.0.) go to 640
c calculate the parameters of the givens transformation.
call fpgivs(piv,av1(j,1),co,si)
c apply that transformation to the columns of matrix g.
ic = j
do 610 i=1,nuu
call fprota(co,si,right(i),c(ic))
ic = ic+nv7
610 continue
c apply that transformation to the rows of (avv) with respect to av2.
do 620 i=1,4
call fprota(co,si,h2(i),av2(j,i))
620 continue
c apply that transformation to the rows of (avv) with respect to av1.
if(i2.eq.0) go to 670
do 630 i=1,i2
i1 = i+1
call fprota(co,si,h1(i1),av1(j,i1))
630 continue
640 do 650 i=1,i2
h1(i) = h1(i+1)
650 continue
h1(i2+1) = 0.
660 continue
c rotations with the rows nv11+1,...,nv7 of avv.
670 do 700 j=1,4
ij = nv11+j
if(ij.le.0) go to 700
piv = h2(j)
if(piv.eq.0.) go to 700
c calculate the parameters of the givens transformation.
call fpgivs(piv,av2(ij,j),co,si)
c apply that transformation to the columns of matrix g.
ic = ij
do 680 i=1,nuu
call fprota(co,si,right(i),c(ic))
ic = ic+nv7
680 continue
if(j.eq.4) go to 700
c apply that transformation to the rows of (avv) with respect to av2.
j1 = j+1
do 690 i=j1,4
call fprota(co,si,h2(i),av2(ij,i))
690 continue
700 continue
c we update the sum of squared residuals
do 705 i=1,nuu
sq = sq+right(i)**2
705 continue
go to 750
c rotation into triangle of the new row of (avv), in case the elements
c corresponding to the b-splines n(j;v),j=nv7+1,...,nv4 are all zero.
710 irot =nrold
do 740 i=1,5
irot = irot+1
piv = h(i)
if(piv.eq.0.) go to 740
c calculate the parameters of the givens transformation.
call fpgivs(piv,av1(irot,1),co,si)
c apply that transformation to the columns of matrix g.
ic = irot
do 720 j=1,nuu
call fprota(co,si,right(j),c(ic))
ic = ic+nv7
720 continue
c apply that transformation to the rows of (avv).
if(i.eq.5) go to 740
i2 = 1
i3 = i+1
do 730 j=i3,5
i2 = i2+1
call fprota(co,si,h(j),av1(irot,i2))
730 continue
740 continue
c we update the sum of squared residuals
do 745 i=1,nuu
sq = sq+right(i)**2
745 continue
750 if(nrold.eq.number) go to 770
760 nrold = nrold+1
go to 450
770 continue
c test whether the b-spline coefficients must be determined.
if(iback.ne.0) return
c backward substitution to obtain the b-spline coefficients as the
c solution of the linear system (rv) (cr) (ru)' = h.
c first step: solve the system (rv) (c1) = h.
k = 1
do 780 i=1,nuu
call fpbacp(av1,av2,c(k),nv7,4,c(k),5,nv)
k = k+nv7
780 continue
c second step: solve the system (cr) (ru)' = (c1).
k = 0
do 800 j=1,nv7
k = k+1
l = k
do 790 i=1,nuu
right(i) = c(l)
l = l+nv7
790 continue
call fpback(au,right,nuu,5,right,nu)
l = k
do 795 i=1,nuu
c(l) = right(i)
l = l+nv7
795 continue
800 continue
c calculate from the conditions (2)-(3)-(4), the remaining b-spline
c coefficients.
ncof = nu4*nv4
i = nv4
j = 0
do 805 l=1,nv4
q(l) = dz1
805 continue
if(iop0.eq.0) go to 815
do 810 l=1,nv4
i = i+1
q(i) = cc(l)
810 continue
815 if(nuu.eq.0) go to 850
do 840 l=1,nuu
ii = i
do 820 k=1,nv7
i = i+1
j = j+1
q(i) = c(j)
820 continue
do 830 k=1,3
ii = ii+1
i = i+1
q(i) = q(ii)
830 continue
840 continue
850 if(iop1.eq.0) go to 870
do 860 l=1,nv4
i = i+1
q(i) = 0.
860 continue
870 do 880 i=1,ncof
c(i) = q(i)
880 continue
c calculate the quantities
c res(i,j) = (z(i,j) - s(u(i),v(j)))**2 , i=1,2,..,mu;j=1,2,..,mv
c fp = sumi=1,mu(sumj=1,mv(res(i,j)))
c fpu(r) = sum''i(sumj=1,mv(res(i,j))) , r=1,2,...,nu-7
c tu(r+3) <= u(i) <= tu(r+4)
c fpv(r) = sumi=1,mu(sum''j(res(i,j))) , r=1,2,...,nv-7
c tv(r+3) <= v(j) <= tv(r+4)
fp = 0.
do 890 i=1,nu
fpu(i) = 0.
890 continue
do 900 i=1,nv
fpv(i) = 0.
900 continue
iz = 0
nroldu = 0
c main loop for the different grid points.
do 950 i1=1,mu
numu = nru(i1)
numu1 = numu+1
nroldv = 0
do 940 i2=1,mv
numv = nrv(i2)
numv1 = numv+1
iz = iz+1
c evaluate s(u,v) at the current grid point by making the sum of the
c cross products of the non-zero b-splines at (u,v), multiplied with
c the appropriate b-spline coefficients.
term = 0.
k1 = numu*nv4+numv
do 920 l1=1,4
k2 = k1
fac = spu(i1,l1)
do 910 l2=1,4
k2 = k2+1
term = term+fac*spv(i2,l2)*c(k2)
910 continue
k1 = k1+nv4
920 continue
c calculate the squared residual at the current grid point.
term = (z(iz)-term)**2
c adjust the different parameters.
fp = fp+term
fpu(numu1) = fpu(numu1)+term
fpv(numv1) = fpv(numv1)+term
fac = term*half
if(numv.eq.nroldv) go to 930
fpv(numv1) = fpv(numv1)-fac
fpv(numv) = fpv(numv)+fac
930 nroldv = numv
if(numu.eq.nroldu) go to 940
fpu(numu1) = fpu(numu1)-fac
fpu(numu) = fpu(numu)+fac
940 continue
nroldu = numu
950 continue
return
end
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