subroutine dspco(ap,n,kpvt,rcond,z) integer n,kpvt(1) double precision ap(1),z(1) double precision rcond c c dspco factors a double precision symmetric matrix stored in c packed form by elimination with symmetric pivoting and estimates c the condition of the matrix. c c if rcond is not needed, dspfa is slightly faster. c to solve a*x = b , follow dspco by dspsl. c to compute inverse(a)*c , follow dspco by dspsl. c to compute inverse(a) , follow dspco by dspdi. c to compute determinant(a) , follow dspco by dspdi. c to compute inertia(a), follow dspco by dspdi. c c on entry c c ap double precision (n*(n+1)/2) c the packed form of a symmetric matrix a . the c columns of the upper triangle are stored sequentially c in a one-dimensional array of length n*(n+1)/2 . c see comments below for details. c c n integer c the order of the matrix a . c c output c c ap a block diagonal matrix and the multipliers which c were used to obtain it stored in packed form. c the factorization can be written a = u*d*trans(u) c where u is a product of permutation and unit c upper triangular matrices , trans(u) is the c transpose of u , and d is block diagonal c with 1 by 1 and 2 by 2 blocks. c c kpvt integer(n) c an integer vector of pivot indices. c c rcond double precision c an estimate of the reciprocal condition of a . c for the system a*x = b , relative perturbations c in a and b of size epsilon may cause c relative perturbations in x of size epsilon/rcond . c if rcond is so small that the logical expression c 1.0 + rcond .eq. 1.0 c is true, then a may be singular to working c precision. in particular, rcond is zero if c exact singularity is detected or the estimate c underflows. c c z double precision(n) c a work vector whose contents are usually unimportant. c if a is close to a singular matrix, then z is c an approximate null vector in the sense that c norm(a*z) = rcond*norm(a)*norm(z) . c c packed storage c c the following program segment will pack the upper c triangle of a symmetric matrix. c c k = 0 c do 20 j = 1, n c do 10 i = 1, j c k = k + 1 c ap(k) = a(i,j) c 10 continue c 20 continue c c linpack. this version dated 08/14/78 . c cleve moler, university of new mexico, argonne national lab. c c subroutines and functions c c linpack dspfa c blas daxpy,ddot,dscal,dasum c fortran dabs,dmax1,iabs,dsign c c internal variables c double precision ak,akm1,bk,bkm1,ddot,denom,ek,t double precision anorm,s,dasum,ynorm integer i,ij,ik,ikm1,ikp1,info,j,jm1,j1 integer k,kk,km1k,km1km1,kp,kps,ks c c c find norm of a using only upper half c j1 = 1 do 30 j = 1, n z(j) = dasum(j,ap(j1),1) ij = j1 j1 = j1 + j jm1 = j - 1 if (jm1 .lt. 1) go to 20 do 10 i = 1, jm1 z(i) = z(i) + dabs(ap(ij)) ij = ij + 1 10 continue 20 continue 30 continue anorm = 0.0d0 do 40 j = 1, n anorm = dmax1(anorm,z(j)) 40 continue c c factor c call dspfa(ap,n,kpvt,info) c c rcond = 1/(norm(a)*(estimate of norm(inverse(a)))) . c estimate = norm(z)/norm(y) where a*z = y and a*y = e . c the components of e are chosen to cause maximum local c growth in the elements of w where u*d*w = e . c the vectors are frequently rescaled to avoid overflow. c c solve u*d*w = e c ek = 1.0d0 do 50 j = 1, n z(j) = 0.0d0 50 continue k = n ik = (n*(n - 1))/2 60 if (k .eq. 0) go to 120 kk = ik + k ikm1 = ik - (k - 1) ks = 1 if (kpvt(k) .lt. 0) ks = 2 kp = iabs(kpvt(k)) kps = k + 1 - ks if (kp .eq. kps) go to 70 t = z(kps) z(kps) = z(kp) z(kp) = t 70 continue if (z(k) .ne. 0.0d0) ek = dsign(ek,z(k)) z(k) = z(k) + ek call daxpy(k-ks,z(k),ap(ik+1),1,z(1),1) if (ks .eq. 1) go to 80 if (z(k-1) .ne. 0.0d0) ek = dsign(ek,z(k-1)) z(k-1) = z(k-1) + ek call daxpy(k-ks,z(k-1),ap(ikm1+1),1,z(1),1) 80 continue if (ks .eq. 2) go to 100 if (dabs(z(k)) .le. dabs(ap(kk))) go to 90 s = dabs(ap(kk))/dabs(z(k)) call dscal(n,s,z,1) ek = s*ek 90 continue if (ap(kk) .ne. 0.0d0) z(k) = z(k)/ap(kk) if (ap(kk) .eq. 0.0d0) z(k) = 1.0d0 go to 110 100 continue km1k = ik + k - 1 km1km1 = ikm1 + k - 1 ak = ap(kk)/ap(km1k) akm1 = ap(km1km1)/ap(km1k) bk = z(k)/ap(km1k) bkm1 = z(k-1)/ap(km1k) denom = ak*akm1 - 1.0d0 z(k) = (akm1*bk - bkm1)/denom z(k-1) = (ak*bkm1 - bk)/denom 110 continue k = k - ks ik = ik - k if (ks .eq. 2) ik = ik - (k + 1) go to 60 120 continue s = 1.0d0/dasum(n,z,1) call dscal(n,s,z,1) c c solve trans(u)*y = w c k = 1 ik = 0 130 if (k .gt. n) go to 160 ks = 1 if (kpvt(k) .lt. 0) ks = 2 if (k .eq. 1) go to 150 z(k) = z(k) + ddot(k-1,ap(ik+1),1,z(1),1) ikp1 = ik + k if (ks .eq. 2) * z(k+1) = z(k+1) + ddot(k-1,ap(ikp1+1),1,z(1),1) kp = iabs(kpvt(k)) if (kp .eq. k) go to 140 t = z(k) z(k) = z(kp) z(kp) = t 140 continue 150 continue ik = ik + k if (ks .eq. 2) ik = ik + (k + 1) k = k + ks go to 130 160 continue s = 1.0d0/dasum(n,z,1) call dscal(n,s,z,1) c ynorm = 1.0d0 c c solve u*d*v = y c k = n ik = n*(n - 1)/2 170 if (k .eq. 0) go to 230 kk = ik + k ikm1 = ik - (k - 1) ks = 1 if (kpvt(k) .lt. 0) ks = 2 if (k .eq. ks) go to 190 kp = iabs(kpvt(k)) kps = k + 1 - ks if (kp .eq. kps) go to 180 t = z(kps) z(kps) = z(kp) z(kp) = t 180 continue call daxpy(k-ks,z(k),ap(ik+1),1,z(1),1) if (ks .eq. 2) call daxpy(k-ks,z(k-1),ap(ikm1+1),1,z(1),1) 190 continue if (ks .eq. 2) go to 210 if (dabs(z(k)) .le. dabs(ap(kk))) go to 200 s = dabs(ap(kk))/dabs(z(k)) call dscal(n,s,z,1) ynorm = s*ynorm 200 continue if (ap(kk) .ne. 0.0d0) z(k) = z(k)/ap(kk) if (ap(kk) .eq. 0.0d0) z(k) = 1.0d0 go to 220 210 continue km1k = ik + k - 1 km1km1 = ikm1 + k - 1 ak = ap(kk)/ap(km1k) akm1 = ap(km1km1)/ap(km1k) bk = z(k)/ap(km1k) bkm1 = z(k-1)/ap(km1k) denom = ak*akm1 - 1.0d0 z(k) = (akm1*bk - bkm1)/denom z(k-1) = (ak*bkm1 - bk)/denom 220 continue k = k - ks ik = ik - k if (ks .eq. 2) ik = ik - (k + 1) go to 170 230 continue s = 1.0d0/dasum(n,z,1) call dscal(n,s,z,1) ynorm = s*ynorm c c solve trans(u)*z = v c k = 1 ik = 0 240 if (k .gt. n) go to 270 ks = 1 if (kpvt(k) .lt. 0) ks = 2 if (k .eq. 1) go to 260 z(k) = z(k) + ddot(k-1,ap(ik+1),1,z(1),1) ikp1 = ik + k if (ks .eq. 2) * z(k+1) = z(k+1) + ddot(k-1,ap(ikp1+1),1,z(1),1) kp = iabs(kpvt(k)) if (kp .eq. k) go to 250 t = z(k) z(k) = z(kp) z(kp) = t 250 continue 260 continue ik = ik + k if (ks .eq. 2) ik = ik + (k + 1) k = k + ks go to 240 270 continue c make znorm = 1.0 s = 1.0d0/dasum(n,z,1) call dscal(n,s,z,1) ynorm = s*ynorm c if (anorm .ne. 0.0d0) rcond = ynorm/anorm if (anorm .eq. 0.0d0) rcond = 0.0d0 return end