1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
|
subroutine dpoco(a,lda,n,rcond,z,info)
integer lda,n,info
double precision a(lda,1),z(1)
double precision rcond
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 dabs,dmax1,dreal,dsign
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) + dabs(a(i,j))
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 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 = dsign(ek,-z(k))
if (dabs(ek-z(k)) .le. a(k,k)) go to 60
s = a(k,k)/dabs(ek-z(k))
call dscal(n,s,z,1)
ek = s*ek
60 continue
wk = ek - z(k)
wkm = -ek - z(k)
s = dabs(wk)
sm = dabs(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 + dabs(z(j)+wkm*a(k,j))
z(j) = z(j) + wk*a(k,j)
s = s + dabs(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 (dabs(z(k)) .le. a(k,k)) go to 120
s = a(k,k)/dabs(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 (dabs(z(k)) .le. a(k,k)) go to 140
s = a(k,k)/dabs(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 (dabs(z(k)) .le. a(k,k)) go to 160
s = a(k,k)/dabs(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
|
