subroutine sspco(ap,n,kpvt,rcond,z)
integer n,kpvt(1)
real ap(1),z(1)
real rcond
c
c sspco factors a real symmetric matrix stored in packed
c form by elimination with symmetric pivoting and estimates
c the condition of the matrix.
c
c if rcond is not needed, sspfa is slightly faster.
c to solve a*x = b , follow sspco by sspsl.
c to compute inverse(a)*c , follow sspco by sspsl.
c to compute inverse(a) , follow sspco by sspdi.
c to compute determinant(a) , follow sspco by sspdi.
c to compute inertia(a), follow sspco by sspdi.
c
c on entry
c
c ap real (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 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.
c
c z real(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 sspfa
c blas saxpy,sdot,sscal,sasum
c fortran abs,amax1,iabs,sign
c
c internal variables
c
real ak,akm1,bk,bkm1,sdot,denom,ek,t
real anorm,s,sasum,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) = sasum(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) + abs(ap(ij))
ij = ij + 1
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 sspfa(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 = 1.0e0
do 50 j = 1, n
z(j) = 0.0e0
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. 0.0e0) ek = sign(ek,z(k))
z(k) = z(k) + ek
call saxpy(k-ks,z(k),ap(ik+1),1,z(1),1)
if (ks .eq. 1) go to 80
if (z(k-1) .ne. 0.0e0) ek = sign(ek,z(k-1))
z(k-1) = z(k-1) + ek
call saxpy(k-ks,z(k-1),ap(ikm1+1),1,z(1),1)
80 continue
if (ks .eq. 2) go to 100
if (abs(z(k)) .le. abs(ap(kk))) go to 90
s = abs(ap(kk))/abs(z(k))
call sscal(n,s,z,1)
ek = s*ek
90 continue
if (ap(kk) .ne. 0.0e0) z(k) = z(k)/ap(kk)
if (ap(kk) .eq. 0.0e0) z(k) = 1.0e0
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 - 1.0e0
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 = 1.0e0/sasum(n,z,1)
call sscal(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) + sdot(k-1,ap(ik+1),1,z(1),1)
ikp1 = ik + k
if (ks .eq. 2)
* z(k+1) = z(k+1) + sdot(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 = 1.0e0/sasum(n,z,1)
call sscal(n,s,z,1)
c
ynorm = 1.0e0
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 saxpy(k-ks,z(k),ap(ik+1),1,z(1),1)
if (ks .eq. 2) call saxpy(k-ks,z(k-1),ap(ikm1+1),1,z(1),1)
190 continue
if (ks .eq. 2) go to 210
if (abs(z(k)) .le. abs(ap(kk))) go to 200
s = abs(ap(kk))/abs(z(k))
call sscal(n,s,z,1)
ynorm = s*ynorm
200 continue
if (ap(kk) .ne. 0.0e0) z(k) = z(k)/ap(kk)
if (ap(kk) .eq. 0.0e0) z(k) = 1.0e0
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 - 1.0e0
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 = 1.0e0/sasum(n,z,1)
call sscal(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) + sdot(k-1,ap(ik+1),1,z(1),1)
ikp1 = ik + k
if (ks .eq. 2)
* z(k+1) = z(k+1) + sdot(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 = 1.0e0/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
return
end
.