subroutine spbco(abd,lda,n,m,rcond,z,info) integer lda,n,m,info real abd(lda,1),z(1) real rcond c c spbco factors a real symmetric positive definite matrix c stored in band form and estimates the condition of the matrix. c c if rcond is not needed, spbfa is slightly faster. c to solve a*x = b , follow spbco by spbsl. c to compute inverse(a)*c , follow spbco by spbsl. c to compute determinant(a) , follow spbco by spbdi. c c on entry c c abd real(lda, n) c the matrix to be factored. the columns of the upper c triangle are stored in the columns of abd and the c diagonals of the upper triangle are stored in the c rows of abd . see the comments below for details. c c lda integer c the leading dimension of the array abd . c lda must be .ge. m + 1 . c c n integer c the order of the matrix a . c c m integer c the number of diagonals above the main diagonal. c 0 .le. m .lt. n . c c on return c c abd an upper triangular matrix r , stored in band c form, so that a = trans(r)*r . c if info .ne. 0 , the factorization is not complete. c c rcond real 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 real(n) c a work vector whose contents are usually unimportant. c if a is singular to working precision, 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 band storage c c if a is a symmetric positive definite band matrix, c the following program segment will set up the input. c c m = (band width above diagonal) c do 20 j = 1, n c i1 = max0(1, j-m) c do 10 i = i1, j c k = i-j+m+1 c abd(k,j) = a(i,j) c 10 continue c 20 continue c c this uses m + 1 rows of a , except for the m by m c upper left triangle, which is ignored. c c example.. if the original matrix is c c 11 12 13 0 0 0 c 12 22 23 24 0 0 c 13 23 33 34 35 0 c 0 24 34 44 45 46 c 0 0 35 45 55 56 c 0 0 0 46 56 66 c c then n = 6 , m = 2 and abd should contain c c * * 13 24 35 46 c * 12 23 34 45 56 c 11 22 33 44 55 66 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 spbfa c blas saxpy,sdot,sscal,sasum c fortran abs,amax1,max0,min0,real,sign c c internal variables c real sdot,ek,t,wk,wkm real anorm,s,sasum,sm,ynorm integer i,j,j2,k,kb,kp1,l,la,lb,lm,mu c c c find norm of a c do 30 j = 1, n l = min0(j,m+1) mu = max0(m+2-j,1) z(j) = sasum(l,abd(mu,j),1) k = j - l if (m .lt. mu) go to 20 do 10 i = mu, m k = k + 1 z(k) = z(k) + abs(abd(i,j)) 10 continue 20 continue 30 continue anorm = 0.0e0 do 40 j = 1, n anorm = amax1(anorm,z(j)) 40 continue c c factor c call spbfa(abd,lda,n,m,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 = 1.0e0 do 50 j = 1, n z(j) = 0.0e0 50 continue do 110 k = 1, n if (z(k) .ne. 0.0e0) ek = sign(ek,-z(k)) if (abs(ek-z(k)) .le. abd(m+1,k)) go to 60 s = abd(m+1,k)/abs(ek-z(k)) call sscal(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/abd(m+1,k) wkm = wkm/abd(m+1,k) kp1 = k + 1 j2 = min0(k+m,n) i = m + 1 if (kp1 .gt. j2) go to 100 do 70 j = kp1, j2 i = i - 1 sm = sm + abs(z(j)+wkm*abd(i,j)) z(j) = z(j) + wk*abd(i,j) s = s + abs(z(j)) 70 continue if (s .ge. sm) go to 90 t = wkm - wk wk = wkm i = m + 1 do 80 j = kp1, j2 i = i - 1 z(j) = z(j) + t*abd(i,j) 80 continue 90 continue 100 continue z(k) = wk 110 continue s = 1.0e0/sasum(n,z,1) call sscal(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. abd(m+1,k)) go to 120 s = abd(m+1,k)/abs(z(k)) call sscal(n,s,z,1) 120 continue z(k) = z(k)/abd(m+1,k) lm = min0(k-1,m) la = m + 1 - lm lb = k - lm t = -z(k) call saxpy(lm,t,abd(la,k),1,z(lb),1) 130 continue s = 1.0e0/sasum(n,z,1) call sscal(n,s,z,1) c ynorm = 1.0e0 c c solve trans(r)*v = y c do 150 k = 1, n lm = min0(k-1,m) la = m + 1 - lm lb = k - lm z(k) = z(k) - sdot(lm,abd(la,k),1,z(lb),1) if (abs(z(k)) .le. abd(m+1,k)) go to 140 s = abd(m+1,k)/abs(z(k)) call sscal(n,s,z,1) ynorm = s*ynorm 140 continue z(k) = z(k)/abd(m+1,k) 150 continue s = 1.0e0/sasum(n,z,1) call sscal(n,s,z,1) ynorm = s*ynorm c c solve r*z = w c do 170 kb = 1, n k = n + 1 - kb if (abs(z(k)) .le. abd(m+1,k)) go to 160 s = abd(m+1,k)/abs(z(k)) call sscal(n,s,z,1) ynorm = s*ynorm 160 continue z(k) = z(k)/abd(m+1,k) lm = min0(k-1,m) la = m + 1 - lm lb = k - lm t = -z(k) call saxpy(lm,t,abd(la,k),1,z(lb),1) 170 continue c make znorm = 1.0 s = 1.0e0/sasum(n,z,1) call sscal(n,s,z,1) ynorm = s*ynorm c if (anorm .ne. 0.0e0) rcond = ynorm/anorm if (anorm .eq. 0.0e0) rcond = 0.0e0 180 continue return end