c c dpoco factors a double precision symmetric positive definite c matrix and estimates the condition of the matrix. c c if rcond is not needed, dpofa is slightly faster. c to solve a*x = b , follow dpoco by dposl. c to compute inverse(a)*c , follow dpoco by dposl. c to compute determinant(a) , follow dpoco by dpodi. c to compute inverse(a) , follow dpoco by dpodi. c c on entry c c a double precision(lda, n) c the symmetric matrix to be factored. only the c diagonal and upper triangle are used. c c lda integer c the leading dimension of the array a . c c n integer c the order of the matrix a . c c on return c c a an upper triangular matrix r so that a = trans(r)*r c where trans(r) is the transpose. c the strict lower triangle is unaltered. c if info .ne. 0 , the factorization is not complete. 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. if info .ne. 0 , rcond is unchanged. 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 if info .ne. 0 , z is unchanged. c c info integer c = 0 for normal return. c = k signals an error condition. the leading minor c of order k is not positive definite. 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 dpofa c blas daxpy,ddot,dscal,dasum c fortran abs,max,sign c subroutine dpoco(a,lda,n,rcond,z,info) integer lda,n,info double precision a(lda,n),z(n) double precision rcond c c internal variables c double precision ddot,ek,t,wk,wkm double precision anorm,s,dasum,sm,ynorm integer i,j,jm1,k,kb,kp1 c c c find norm of a using only upper half c do 30 j = 1, n z(j) = dasum(j,a(1,j),1) jm1 = j - 1 if (jm1 .lt. 1) go to 20 do 10 i = 1, jm1 z(i) = z(i) + abs(a(i,j)) 10 continue 20 continue 30 continue anorm = 0.0d0 do 40 j = 1, n anorm = max(anorm,z(j)) 40 continue c c factor c call dpofa(a,lda,n,info) if (info .ne. 0) go to 180 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 trans(r)*w = e . c the vectors are frequently rescaled to avoid overflow. c c solve trans(r)*w = e c ek = 1.0d0 do 50 j = 1, n z(j) = 0.0d0 50 continue do 110 k = 1, n if (z(k) .ne. 0.0d0) ek = sign(ek,-z(k)) if (abs(ek-z(k)) .le. a(k,k)) go to 60 s = a(k,k)/abs(ek-z(k)) call dscal(n,s,z,1) ek = s*ek 60 continue wk = ek - z(k) wkm = -ek - z(k) s = abs(wk) sm = abs(wkm) wk = wk/a(k,k) wkm = wkm/a(k,k) kp1 = k + 1 if (kp1 .gt. n) go to 100 do 70 j = kp1, n sm = sm + abs(z(j)+wkm*a(k,j)) z(j) = z(j) + wk*a(k,j) s = s + abs(z(j)) 70 continue if (s .ge. sm) go to 90 t = wkm - wk wk = wkm do 80 j = kp1, n z(j) = z(j) + t*a(k,j) 80 continue 90 continue 100 continue z(k) = wk 110 continue s = 1.0d0/dasum(n,z,1) call dscal(n,s,z,1) c c solve r*y = w c do 130 kb = 1, n k = n + 1 - kb if (abs(z(k)) .le. a(k,k)) go to 120 s = a(k,k)/abs(z(k)) call dscal(n,s,z,1) 120 continue z(k) = z(k)/a(k,k) t = -z(k) call daxpy(k-1,t,a(1,k),1,z(1),1) 130 continue s = 1.0d0/dasum(n,z,1) call dscal(n,s,z,1) c ynorm = 1.0d0 c c solve trans(r)*v = y c do 150 k = 1, n z(k) = z(k) - ddot(k-1,a(1,k),1,z(1),1) if (abs(z(k)) .le. a(k,k)) go to 140 s = a(k,k)/abs(z(k)) call dscal(n,s,z,1) ynorm = s*ynorm 140 continue z(k) = z(k)/a(k,k) 150 continue s = 1.0d0/dasum(n,z,1) call dscal(n,s,z,1) ynorm = s*ynorm c c solve r*z = v c do 170 kb = 1, n k = n + 1 - kb if (abs(z(k)) .le. a(k,k)) go to 160 s = a(k,k)/abs(z(k)) call dscal(n,s,z,1) ynorm = s*ynorm 160 continue z(k) = z(k)/a(k,k) t = -z(k) call daxpy(k-1,t,a(1,k),1,z(1),1) 170 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 180 continue return end