-
Notifications
You must be signed in to change notification settings - Fork 0
/
Copy pathLINVERSE.SIF
281 lines (207 loc) · 7.06 KB
/
LINVERSE.SIF
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
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
***************************
* SET UP THE INITIAL DATA *
***************************
NAME LINVERSE
* Problem :
* *********
* The problem is to find the positive definite lower bidiagonal
* matrix L such that the matrix L(inv)L(inv-transp) best approximates,
* in the Frobenius norm, a given symmetric target matrix T.
* More precisely, one is interested in the positive definite lower
* bidiagonal L such that
* || L T L(transp) - I || is minimum.
* F
* The positive definite character of L is imposed by requiring
* that all its diagonal entries to be at least equal to EPSILON,
* a strictly positive real number.
* Many variants of the problem can be obtained by varying the target
* matrix T and the scalar EPSILON. In the present problem,
* a) T is chosen to be pentadiagonal with T(i,j) = sin(i)cos(j) (j .leq. i)
* b) EPSILON = 1.D-8
* Source:
* Ph. Toint, private communication, 1991.
* SIF input: Ph. Toint, March 1991.
* classification SBR2-AN-V-0
* Dimension of the matrix
*IE N 10 $-PARAMETER n = 19 original value
*IE N 100 $-PARAMETER n = 199
*IE N 500 $-PARAMETER n = 999
IE N 1000 $-PARAMETER n = 1999
* Positive definiteness treshold
RE EPSILON 1.0D-8
* Constants
IE 1 1
IE 2 2
IE 3 3
IE 4 4
IA N-1 N -1
IA N-2 N -2
* Target matrix
DO J 1 N-2
IA J+2 J 2
RI RJ J
R( COSJ COS RJ
DO I J J+2
RI RI I
R( SINI SIN RI
A* T(I,J) SINI COSJ
ND
RI RN-1 N-1
R( SINI SIN RN-1
R( COSJ COS RN-1
A* T(N-1,N-1)SINI COSJ
RI RN N
R( SINI SIN RN
A* T(N,N-1) SINI COSJ
R( COSJ COS RN
A* T(N,N) SINI COSJ
VARIABLES
DO I 1 N-1
X A(I)
X B(I)
ND
X A(N)
GROUPS
DO J 1 N-2
IA J+1 J 1
IA J+2 J 2
XN O(J,J)
XN O(J+1,J) 'SCALE' 0.5
XN O(J+2,J) 'SCALE' 0.5
ND
XN O(N-1,N-1)
XN O(N,N-1) 'SCALE' 0.5
XN O(N,N)
CONSTANTS
DO I 1 N
X LINVERSE O(I,I) 1.0
ND
BOUNDS
FR LINVERSE 'DEFAULT'
DO I 1 N
ZL LINVERSE A(I) EPSILON
ND
START POINT
XV LINVERSE 'DEFAULT' -1.0
ELEMENT TYPE
EV 2PR X Y
ELEMENT USES
XT 'DEFAULT' 2PR
ZV S(1,1) X A(1)
ZV S(1,1) Y A(1)
ZV S(2,1) X A(2)
ZV S(2,1) Y A(1)
ZV V(2,1) X B(1)
ZV V(2,1) Y A(1)
ZV S(3,1) X A(3)
ZV S(3,1) Y A(1)
ZV V(3,1) X B(2)
ZV V(3,1) Y A(1)
ZV S(2,2) X A(2)
ZV S(2,2) Y A(2)
ZV U(2,2) X A(2)
ZV U(2,2) Y B(1)
ZV V(2,2) X A(2)
ZV V(2,2) Y B(1)
ZV W(2,2) X B(1)
ZV W(2,2) Y B(1)
ZV S(3,2) X A(3)
ZV S(3,2) Y A(2)
ZV U(3,2) X A(3)
ZV U(3,2) Y B(1)
ZV V(3,2) X A(2)
ZV V(3,2) Y B(2)
ZV W(3,2) X B(2)
ZV W(3,2) Y B(1)
ZV S(3,3) X A(3)
ZV S(3,3) Y A(3)
ZV U(3,3) X A(3)
ZV U(3,3) Y B(2)
ZV V(3,3) X A(3)
ZV V(3,3) Y B(2)
ZV W(3,3) X B(2)
ZV W(3,3) Y B(2)
DO I 4 N
IA I-1 I -1
IA I-2 I -2
ZV S(I,I-2) X A(I)
ZV S(I,I-2) Y A(I-2)
ZV V(I,I-2) X B(I-1)
ZV V(I,I-2) Y A(I-2)
DO J I-1 I
IA J-1 J -1
ZV S(I,J) X A(I)
ZV S(I,J) Y A(J)
ZV U(I,J) X A(I)
ZV U(I,J) Y B(J-1)
ZV V(I,J) X B(I-1)
ZV V(I,J) Y A(J)
ZV W(I,J) X B(I-1)
ZV W(I,J) Y B(J-1)
ND
GROUP TYPE
GV L2 GVAR
GROUP USES
XT 'DEFAULT' L2
ZE O(1,1) S(1,1) T(1,1)
ZE O(2,1) S(2,1) T(2,1)
ZE O(2,1) V(2,1) T(1,1)
ZE O(3,1) S(3,1) T(3,1)
ZE O(3,1) V(3,1) T(2,1)
ZE O(2,2) S(2,2) T(2,2)
ZE O(2,2) U(2,2) T(2,1)
ZE O(2,2) V(2,2) T(2,1)
ZE O(2,2) W(2,2) T(1,1)
ZE O(3,2) S(3,2) T(3,2)
ZE O(3,2) U(3,2) T(3,1)
ZE O(3,2) V(3,2) T(2,2)
ZE O(3,2) W(3,2) T(2,1)
ZE O(3,3) S(3,3) T(3,3)
ZE O(3,3) U(3,3) T(3,2)
ZE O(3,3) V(3,3) T(3,2)
ZE O(3,3) W(3,3) T(2,2)
DO I 4 N
IA I-1 I -1
IA I-2 I -2
ZE O(I,I-2) S(I,I-2) T(I,I-2)
ZE O(I,I-2) V(I,I-2) T(I-1,I-2)
ZE O(I,I-1) S(I,I-1) T(I,I-1)
ZE O(I,I-1) U(I,I-1) T(I,I-2)
ZE O(I,I-1) V(I,I-1) T(I-1,I-1)
ZE O(I,I-1) W(I,I-1) T(I-1,I-2)
ZE O(I,I) S(I,I) T(I,I)
ZE O(I,I) U(I,I) T(I,I-1)
ZE O(I,I) V(I,I) T(I,I-1)
ZE O(I,I) W(I,I) T(I-1,I-1)
ND
OBJECT BOUND
* Solution
*LO SOLTN(10) 6.00000000
*LO SOLTN(100) 68.0000000
*LO SOLTN(500) 340.000000
*LO SOLTN(1000) ???
ENDATA
***********************
* SET UP THE FUNCTION *
* AND RANGE ROUTINES *
***********************
ELEMENTS LINVERSE
INDIVIDUALS
T 2PR
F X * Y
G X Y
G Y X
H X Y 1.0
ENDATA
*********************
* SET UP THE GROUPS *
* ROUTINE *
*********************
GROUPS LINVERSE
* Least-square groups
INDIVIDUALS
T L2
F GVAR * GVAR
G GVAR + GVAR
H 2.0
ENDATA