| 1 | /*************************************************************************
|
|---|
| 2 | Copyright (c) 1992-2007 The University of Tennessee. All rights reserved.
|
|---|
| 3 |
|
|---|
| 4 | Contributors:
|
|---|
| 5 | * Sergey Bochkanov (ALGLIB project). Translation from FORTRAN to
|
|---|
| 6 | pseudocode.
|
|---|
| 7 |
|
|---|
| 8 | See subroutines comments for additional copyrights.
|
|---|
| 9 |
|
|---|
| 10 | >>> SOURCE LICENSE >>>
|
|---|
| 11 | This program is free software; you can redistribute it and/or modify
|
|---|
| 12 | it under the terms of the GNU General Public License as published by
|
|---|
| 13 | the Free Software Foundation (www.fsf.org); either version 2 of the
|
|---|
| 14 | License, or (at your option) any later version.
|
|---|
| 15 |
|
|---|
| 16 | This program is distributed in the hope that it will be useful,
|
|---|
| 17 | but WITHOUT ANY WARRANTY; without even the implied warranty of
|
|---|
| 18 | MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
|
|---|
| 19 | GNU General Public License for more details.
|
|---|
| 20 |
|
|---|
| 21 | A copy of the GNU General Public License is available at
|
|---|
| 22 | http://www.fsf.org/licensing/licenses
|
|---|
| 23 |
|
|---|
| 24 | >>> END OF LICENSE >>>
|
|---|
| 25 | *************************************************************************/
|
|---|
| 26 |
|
|---|
| 27 | using System;
|
|---|
| 28 |
|
|---|
| 29 | namespace alglib
|
|---|
| 30 | {
|
|---|
| 31 | public class srcond
|
|---|
| 32 | {
|
|---|
| 33 | /*************************************************************************
|
|---|
| 34 | Condition number estimate of a symmetric matrix
|
|---|
| 35 |
|
|---|
| 36 | The algorithm calculates a lower bound of the condition number. In this
|
|---|
| 37 | case, the algorithm does not return a lower bound of the condition number,
|
|---|
| 38 | but an inverse number (to avoid an overflow in case of a singular matrix).
|
|---|
| 39 |
|
|---|
| 40 | It should be noted that 1-norm and inf-norm condition numbers of symmetric
|
|---|
| 41 | matrices are equal, so the algorithm doesn't take into account the
|
|---|
| 42 | differences between these types of norms.
|
|---|
| 43 |
|
|---|
| 44 | Input parameters:
|
|---|
| 45 | A - symmetric definite matrix which is given by its upper or
|
|---|
| 46 | lower triangle depending on IsUpper.
|
|---|
| 47 | Array with elements [0..N-1, 0..N-1].
|
|---|
| 48 | N - size of matrix A.
|
|---|
| 49 | IsUpper - storage format.
|
|---|
| 50 |
|
|---|
| 51 | Result:
|
|---|
| 52 | 1/LowerBound(cond(A))
|
|---|
| 53 | *************************************************************************/
|
|---|
| 54 | public static double smatrixrcond(ref double[,] a,
|
|---|
| 55 | int n,
|
|---|
| 56 | bool isupper)
|
|---|
| 57 | {
|
|---|
| 58 | double result = 0;
|
|---|
| 59 | int i = 0;
|
|---|
| 60 | int j = 0;
|
|---|
| 61 | double[,] a1 = new double[0,0];
|
|---|
| 62 |
|
|---|
| 63 | a1 = new double[n+1, n+1];
|
|---|
| 64 | for(i=1; i<=n; i++)
|
|---|
| 65 | {
|
|---|
| 66 | if( isupper )
|
|---|
| 67 | {
|
|---|
| 68 | for(j=i; j<=n; j++)
|
|---|
| 69 | {
|
|---|
| 70 | a1[i,j] = a[i-1,j-1];
|
|---|
| 71 | }
|
|---|
| 72 | }
|
|---|
| 73 | else
|
|---|
| 74 | {
|
|---|
| 75 | for(j=1; j<=i; j++)
|
|---|
| 76 | {
|
|---|
| 77 | a1[i,j] = a[i-1,j-1];
|
|---|
| 78 | }
|
|---|
| 79 | }
|
|---|
| 80 | }
|
|---|
| 81 | result = rcondsymmetric(a1, n, isupper);
|
|---|
| 82 | return result;
|
|---|
| 83 | }
|
|---|
| 84 |
|
|---|
| 85 |
|
|---|
| 86 | /*************************************************************************
|
|---|
| 87 | Condition number estimate of a matrix given by LDLT-decomposition
|
|---|
| 88 |
|
|---|
| 89 | The algorithm calculates a lower bound of the condition number. In this
|
|---|
| 90 | case, the algorithm does not return a lower bound of the condition number,
|
|---|
| 91 | but an inverse number (to avoid an overflow in case of a singular matrix).
|
|---|
| 92 |
|
|---|
| 93 | It should be noted that 1-norm and inf-norm condition numbers of symmetric
|
|---|
| 94 | matrices are equal, so the algorithm doesn't take into account the
|
|---|
| 95 | differences between these types of norms.
|
|---|
| 96 |
|
|---|
| 97 | Input parameters:
|
|---|
| 98 | L - LDLT-decomposition of matrix A given by the upper or lower
|
|---|
| 99 | triangle depending on IsUpper.
|
|---|
| 100 | Output of SMatrixLDLT subroutine.
|
|---|
| 101 | Pivots - table of permutations which were made during LDLT-decomposition,
|
|---|
| 102 | Output of SMatrixLDLT subroutine.
|
|---|
| 103 | N - size of matrix A.
|
|---|
| 104 | IsUpper - storage format.
|
|---|
| 105 |
|
|---|
| 106 | Result:
|
|---|
| 107 | 1/LowerBound(cond(A))
|
|---|
| 108 | *************************************************************************/
|
|---|
| 109 | public static double smatrixldltrcond(ref double[,] l,
|
|---|
| 110 | ref int[] pivots,
|
|---|
| 111 | int n,
|
|---|
| 112 | bool isupper)
|
|---|
| 113 | {
|
|---|
| 114 | double result = 0;
|
|---|
| 115 | int i = 0;
|
|---|
| 116 | int j = 0;
|
|---|
| 117 | double[,] l1 = new double[0,0];
|
|---|
| 118 | int[] p1 = new int[0];
|
|---|
| 119 |
|
|---|
| 120 | l1 = new double[n+1, n+1];
|
|---|
| 121 | for(i=1; i<=n; i++)
|
|---|
| 122 | {
|
|---|
| 123 | if( isupper )
|
|---|
| 124 | {
|
|---|
| 125 | for(j=i; j<=n; j++)
|
|---|
| 126 | {
|
|---|
| 127 | l1[i,j] = l[i-1,j-1];
|
|---|
| 128 | }
|
|---|
| 129 | }
|
|---|
| 130 | else
|
|---|
| 131 | {
|
|---|
| 132 | for(j=1; j<=i; j++)
|
|---|
| 133 | {
|
|---|
| 134 | l1[i,j] = l[i-1,j-1];
|
|---|
| 135 | }
|
|---|
| 136 | }
|
|---|
| 137 | }
|
|---|
| 138 | p1 = new int[n+1];
|
|---|
| 139 | for(i=1; i<=n; i++)
|
|---|
| 140 | {
|
|---|
| 141 | if( pivots[i-1]>=0 )
|
|---|
| 142 | {
|
|---|
| 143 | p1[i] = pivots[i-1]+1;
|
|---|
| 144 | }
|
|---|
| 145 | else
|
|---|
| 146 | {
|
|---|
| 147 | p1[i] = -(pivots[i-1]+n+1);
|
|---|
| 148 | }
|
|---|
| 149 | }
|
|---|
| 150 | result = rcondldlt(ref l1, ref p1, n, isupper);
|
|---|
| 151 | return result;
|
|---|
| 152 | }
|
|---|
| 153 |
|
|---|
| 154 |
|
|---|
| 155 | public static double rcondsymmetric(double[,] a,
|
|---|
| 156 | int n,
|
|---|
| 157 | bool isupper)
|
|---|
| 158 | {
|
|---|
| 159 | double result = 0;
|
|---|
| 160 | int i = 0;
|
|---|
| 161 | int j = 0;
|
|---|
| 162 | int im = 0;
|
|---|
| 163 | int jm = 0;
|
|---|
| 164 | double v = 0;
|
|---|
| 165 | double nrm = 0;
|
|---|
| 166 | int[] pivots = new int[0];
|
|---|
| 167 |
|
|---|
| 168 | a = (double[,])a.Clone();
|
|---|
| 169 |
|
|---|
| 170 | nrm = 0;
|
|---|
| 171 | for(j=1; j<=n; j++)
|
|---|
| 172 | {
|
|---|
| 173 | v = 0;
|
|---|
| 174 | for(i=1; i<=n; i++)
|
|---|
| 175 | {
|
|---|
| 176 | im = i;
|
|---|
| 177 | jm = j;
|
|---|
| 178 | if( isupper & j<i )
|
|---|
| 179 | {
|
|---|
| 180 | im = j;
|
|---|
| 181 | jm = i;
|
|---|
| 182 | }
|
|---|
| 183 | if( !isupper & j>i )
|
|---|
| 184 | {
|
|---|
| 185 | im = j;
|
|---|
| 186 | jm = i;
|
|---|
| 187 | }
|
|---|
| 188 | v = v+Math.Abs(a[im,jm]);
|
|---|
| 189 | }
|
|---|
| 190 | nrm = Math.Max(nrm, v);
|
|---|
| 191 | }
|
|---|
| 192 | ldlt.ldltdecomposition(ref a, n, isupper, ref pivots);
|
|---|
| 193 | internalldltrcond(ref a, ref pivots, n, isupper, true, nrm, ref v);
|
|---|
| 194 | result = v;
|
|---|
| 195 | return result;
|
|---|
| 196 | }
|
|---|
| 197 |
|
|---|
| 198 |
|
|---|
| 199 | public static double rcondldlt(ref double[,] l,
|
|---|
| 200 | ref int[] pivots,
|
|---|
| 201 | int n,
|
|---|
| 202 | bool isupper)
|
|---|
| 203 | {
|
|---|
| 204 | double result = 0;
|
|---|
| 205 | double v = 0;
|
|---|
| 206 |
|
|---|
| 207 | internalldltrcond(ref l, ref pivots, n, isupper, false, 0, ref v);
|
|---|
| 208 | result = v;
|
|---|
| 209 | return result;
|
|---|
| 210 | }
|
|---|
| 211 |
|
|---|
| 212 |
|
|---|
| 213 | public static void internalldltrcond(ref double[,] l,
|
|---|
| 214 | ref int[] pivots,
|
|---|
| 215 | int n,
|
|---|
| 216 | bool isupper,
|
|---|
| 217 | bool isnormprovided,
|
|---|
| 218 | double anorm,
|
|---|
| 219 | ref double rcond)
|
|---|
| 220 | {
|
|---|
| 221 | int i = 0;
|
|---|
| 222 | int kase = 0;
|
|---|
| 223 | int k = 0;
|
|---|
| 224 | int km1 = 0;
|
|---|
| 225 | int km2 = 0;
|
|---|
| 226 | int kp1 = 0;
|
|---|
| 227 | int kp2 = 0;
|
|---|
| 228 | double ainvnm = 0;
|
|---|
| 229 | double[] work0 = new double[0];
|
|---|
| 230 | double[] work1 = new double[0];
|
|---|
| 231 | double[] work2 = new double[0];
|
|---|
| 232 | int[] iwork = new int[0];
|
|---|
| 233 | double v = 0;
|
|---|
| 234 | int i_ = 0;
|
|---|
| 235 |
|
|---|
| 236 | System.Diagnostics.Debug.Assert(n>=0);
|
|---|
| 237 |
|
|---|
| 238 | //
|
|---|
| 239 | // Check that the diagonal matrix D is nonsingular.
|
|---|
| 240 | //
|
|---|
| 241 | rcond = 0;
|
|---|
| 242 | if( isupper )
|
|---|
| 243 | {
|
|---|
| 244 | for(i=n; i>=1; i--)
|
|---|
| 245 | {
|
|---|
| 246 | if( pivots[i]>0 & (double)(l[i,i])==(double)(0) )
|
|---|
| 247 | {
|
|---|
| 248 | return;
|
|---|
| 249 | }
|
|---|
| 250 | }
|
|---|
| 251 | }
|
|---|
| 252 | else
|
|---|
| 253 | {
|
|---|
| 254 | for(i=1; i<=n; i++)
|
|---|
| 255 | {
|
|---|
| 256 | if( pivots[i]>0 & (double)(l[i,i])==(double)(0) )
|
|---|
| 257 | {
|
|---|
| 258 | return;
|
|---|
| 259 | }
|
|---|
| 260 | }
|
|---|
| 261 | }
|
|---|
| 262 |
|
|---|
| 263 | //
|
|---|
| 264 | // Estimate the norm of A.
|
|---|
| 265 | //
|
|---|
| 266 | if( !isnormprovided )
|
|---|
| 267 | {
|
|---|
| 268 | kase = 0;
|
|---|
| 269 | anorm = 0;
|
|---|
| 270 | while( true )
|
|---|
| 271 | {
|
|---|
| 272 | estnorm.iterativeestimate1norm(n, ref work1, ref work0, ref iwork, ref anorm, ref kase);
|
|---|
| 273 | if( kase==0 )
|
|---|
| 274 | {
|
|---|
| 275 | break;
|
|---|
| 276 | }
|
|---|
| 277 | if( isupper )
|
|---|
| 278 | {
|
|---|
| 279 |
|
|---|
| 280 | //
|
|---|
| 281 | // Multiply by U'
|
|---|
| 282 | //
|
|---|
| 283 | k = n;
|
|---|
| 284 | while( k>=1 )
|
|---|
| 285 | {
|
|---|
| 286 | if( pivots[k]>0 )
|
|---|
| 287 | {
|
|---|
| 288 |
|
|---|
| 289 | //
|
|---|
| 290 | // P(k)
|
|---|
| 291 | //
|
|---|
| 292 | v = work0[k];
|
|---|
| 293 | work0[k] = work0[pivots[k]];
|
|---|
| 294 | work0[pivots[k]] = v;
|
|---|
| 295 |
|
|---|
| 296 | //
|
|---|
| 297 | // U(k)
|
|---|
| 298 | //
|
|---|
| 299 | km1 = k-1;
|
|---|
| 300 | v = 0.0;
|
|---|
| 301 | for(i_=1; i_<=km1;i_++)
|
|---|
| 302 | {
|
|---|
| 303 | v += work0[i_]*l[i_,k];
|
|---|
| 304 | }
|
|---|
| 305 | work0[k] = work0[k]+v;
|
|---|
| 306 |
|
|---|
| 307 | //
|
|---|
| 308 | // Next k
|
|---|
| 309 | //
|
|---|
| 310 | k = k-1;
|
|---|
| 311 | }
|
|---|
| 312 | else
|
|---|
| 313 | {
|
|---|
| 314 |
|
|---|
| 315 | //
|
|---|
| 316 | // P(k)
|
|---|
| 317 | //
|
|---|
| 318 | v = work0[k-1];
|
|---|
| 319 | work0[k-1] = work0[-pivots[k-1]];
|
|---|
| 320 | work0[-pivots[k-1]] = v;
|
|---|
| 321 |
|
|---|
| 322 | //
|
|---|
| 323 | // U(k)
|
|---|
| 324 | //
|
|---|
| 325 | km1 = k-1;
|
|---|
| 326 | km2 = k-2;
|
|---|
| 327 | v = 0.0;
|
|---|
| 328 | for(i_=1; i_<=km2;i_++)
|
|---|
| 329 | {
|
|---|
| 330 | v += work0[i_]*l[i_,km1];
|
|---|
| 331 | }
|
|---|
| 332 | work0[km1] = work0[km1]+v;
|
|---|
| 333 | v = 0.0;
|
|---|
| 334 | for(i_=1; i_<=km2;i_++)
|
|---|
| 335 | {
|
|---|
| 336 | v += work0[i_]*l[i_,k];
|
|---|
| 337 | }
|
|---|
| 338 | work0[k] = work0[k]+v;
|
|---|
| 339 |
|
|---|
| 340 | //
|
|---|
| 341 | // Next k
|
|---|
| 342 | //
|
|---|
| 343 | k = k-2;
|
|---|
| 344 | }
|
|---|
| 345 | }
|
|---|
| 346 |
|
|---|
| 347 | //
|
|---|
| 348 | // Multiply by D
|
|---|
| 349 | //
|
|---|
| 350 | k = n;
|
|---|
| 351 | while( k>=1 )
|
|---|
| 352 | {
|
|---|
| 353 | if( pivots[k]>0 )
|
|---|
| 354 | {
|
|---|
| 355 | work0[k] = work0[k]*l[k,k];
|
|---|
| 356 | k = k-1;
|
|---|
| 357 | }
|
|---|
| 358 | else
|
|---|
| 359 | {
|
|---|
| 360 | v = work0[k-1];
|
|---|
| 361 | work0[k-1] = l[k-1,k-1]*work0[k-1]+l[k-1,k]*work0[k];
|
|---|
| 362 | work0[k] = l[k-1,k]*v+l[k,k]*work0[k];
|
|---|
| 363 | k = k-2;
|
|---|
| 364 | }
|
|---|
| 365 | }
|
|---|
| 366 |
|
|---|
| 367 | //
|
|---|
| 368 | // Multiply by U
|
|---|
| 369 | //
|
|---|
| 370 | k = 1;
|
|---|
| 371 | while( k<=n )
|
|---|
| 372 | {
|
|---|
| 373 | if( pivots[k]>0 )
|
|---|
| 374 | {
|
|---|
| 375 |
|
|---|
| 376 | //
|
|---|
| 377 | // U(k)
|
|---|
| 378 | //
|
|---|
| 379 | km1 = k-1;
|
|---|
| 380 | v = work0[k];
|
|---|
| 381 | for(i_=1; i_<=km1;i_++)
|
|---|
| 382 | {
|
|---|
| 383 | work0[i_] = work0[i_] + v*l[i_,k];
|
|---|
| 384 | }
|
|---|
| 385 |
|
|---|
| 386 | //
|
|---|
| 387 | // P(k)
|
|---|
| 388 | //
|
|---|
| 389 | v = work0[k];
|
|---|
| 390 | work0[k] = work0[pivots[k]];
|
|---|
| 391 | work0[pivots[k]] = v;
|
|---|
| 392 |
|
|---|
| 393 | //
|
|---|
| 394 | // Next k
|
|---|
| 395 | //
|
|---|
| 396 | k = k+1;
|
|---|
| 397 | }
|
|---|
| 398 | else
|
|---|
| 399 | {
|
|---|
| 400 |
|
|---|
| 401 | //
|
|---|
| 402 | // U(k)
|
|---|
| 403 | //
|
|---|
| 404 | km1 = k-1;
|
|---|
| 405 | kp1 = k+1;
|
|---|
| 406 | v = work0[k];
|
|---|
| 407 | for(i_=1; i_<=km1;i_++)
|
|---|
| 408 | {
|
|---|
| 409 | work0[i_] = work0[i_] + v*l[i_,k];
|
|---|
| 410 | }
|
|---|
| 411 | v = work0[kp1];
|
|---|
| 412 | for(i_=1; i_<=km1;i_++)
|
|---|
| 413 | {
|
|---|
| 414 | work0[i_] = work0[i_] + v*l[i_,kp1];
|
|---|
| 415 | }
|
|---|
| 416 |
|
|---|
| 417 | //
|
|---|
| 418 | // P(k)
|
|---|
| 419 | //
|
|---|
| 420 | v = work0[k];
|
|---|
| 421 | work0[k] = work0[-pivots[k]];
|
|---|
| 422 | work0[-pivots[k]] = v;
|
|---|
| 423 |
|
|---|
| 424 | //
|
|---|
| 425 | // Next k
|
|---|
| 426 | //
|
|---|
| 427 | k = k+2;
|
|---|
| 428 | }
|
|---|
| 429 | }
|
|---|
| 430 | }
|
|---|
| 431 | else
|
|---|
| 432 | {
|
|---|
| 433 |
|
|---|
| 434 | //
|
|---|
| 435 | // Multiply by L'
|
|---|
| 436 | //
|
|---|
| 437 | k = 1;
|
|---|
| 438 | while( k<=n )
|
|---|
| 439 | {
|
|---|
| 440 | if( pivots[k]>0 )
|
|---|
| 441 | {
|
|---|
| 442 |
|
|---|
| 443 | //
|
|---|
| 444 | // P(k)
|
|---|
| 445 | //
|
|---|
| 446 | v = work0[k];
|
|---|
| 447 | work0[k] = work0[pivots[k]];
|
|---|
| 448 | work0[pivots[k]] = v;
|
|---|
| 449 |
|
|---|
| 450 | //
|
|---|
| 451 | // L(k)
|
|---|
| 452 | //
|
|---|
| 453 | kp1 = k+1;
|
|---|
| 454 | v = 0.0;
|
|---|
| 455 | for(i_=kp1; i_<=n;i_++)
|
|---|
| 456 | {
|
|---|
| 457 | v += work0[i_]*l[i_,k];
|
|---|
| 458 | }
|
|---|
| 459 | work0[k] = work0[k]+v;
|
|---|
| 460 |
|
|---|
| 461 | //
|
|---|
| 462 | // Next k
|
|---|
| 463 | //
|
|---|
| 464 | k = k+1;
|
|---|
| 465 | }
|
|---|
| 466 | else
|
|---|
| 467 | {
|
|---|
| 468 |
|
|---|
| 469 | //
|
|---|
| 470 | // P(k)
|
|---|
| 471 | //
|
|---|
| 472 | v = work0[k+1];
|
|---|
| 473 | work0[k+1] = work0[-pivots[k+1]];
|
|---|
| 474 | work0[-pivots[k+1]] = v;
|
|---|
| 475 |
|
|---|
| 476 | //
|
|---|
| 477 | // L(k)
|
|---|
| 478 | //
|
|---|
| 479 | kp1 = k+1;
|
|---|
| 480 | kp2 = k+2;
|
|---|
| 481 | v = 0.0;
|
|---|
| 482 | for(i_=kp2; i_<=n;i_++)
|
|---|
| 483 | {
|
|---|
| 484 | v += work0[i_]*l[i_,k];
|
|---|
| 485 | }
|
|---|
| 486 | work0[k] = work0[k]+v;
|
|---|
| 487 | v = 0.0;
|
|---|
| 488 | for(i_=kp2; i_<=n;i_++)
|
|---|
| 489 | {
|
|---|
| 490 | v += work0[i_]*l[i_,kp1];
|
|---|
| 491 | }
|
|---|
| 492 | work0[kp1] = work0[kp1]+v;
|
|---|
| 493 |
|
|---|
| 494 | //
|
|---|
| 495 | // Next k
|
|---|
| 496 | //
|
|---|
| 497 | k = k+2;
|
|---|
| 498 | }
|
|---|
| 499 | }
|
|---|
| 500 |
|
|---|
| 501 | //
|
|---|
| 502 | // Multiply by D
|
|---|
| 503 | //
|
|---|
| 504 | k = n;
|
|---|
| 505 | while( k>=1 )
|
|---|
| 506 | {
|
|---|
| 507 | if( pivots[k]>0 )
|
|---|
| 508 | {
|
|---|
| 509 | work0[k] = work0[k]*l[k,k];
|
|---|
| 510 | k = k-1;
|
|---|
| 511 | }
|
|---|
| 512 | else
|
|---|
| 513 | {
|
|---|
| 514 | v = work0[k-1];
|
|---|
| 515 | work0[k-1] = l[k-1,k-1]*work0[k-1]+l[k,k-1]*work0[k];
|
|---|
| 516 | work0[k] = l[k,k-1]*v+l[k,k]*work0[k];
|
|---|
| 517 | k = k-2;
|
|---|
| 518 | }
|
|---|
| 519 | }
|
|---|
| 520 |
|
|---|
| 521 | //
|
|---|
| 522 | // Multiply by L
|
|---|
| 523 | //
|
|---|
| 524 | k = n;
|
|---|
| 525 | while( k>=1 )
|
|---|
| 526 | {
|
|---|
| 527 | if( pivots[k]>0 )
|
|---|
| 528 | {
|
|---|
| 529 |
|
|---|
| 530 | //
|
|---|
| 531 | // L(k)
|
|---|
| 532 | //
|
|---|
| 533 | kp1 = k+1;
|
|---|
| 534 | v = work0[k];
|
|---|
| 535 | for(i_=kp1; i_<=n;i_++)
|
|---|
| 536 | {
|
|---|
| 537 | work0[i_] = work0[i_] + v*l[i_,k];
|
|---|
| 538 | }
|
|---|
| 539 |
|
|---|
| 540 | //
|
|---|
| 541 | // P(k)
|
|---|
| 542 | //
|
|---|
| 543 | v = work0[k];
|
|---|
| 544 | work0[k] = work0[pivots[k]];
|
|---|
| 545 | work0[pivots[k]] = v;
|
|---|
| 546 |
|
|---|
| 547 | //
|
|---|
| 548 | // Next k
|
|---|
| 549 | //
|
|---|
| 550 | k = k-1;
|
|---|
| 551 | }
|
|---|
| 552 | else
|
|---|
| 553 | {
|
|---|
| 554 |
|
|---|
| 555 | //
|
|---|
| 556 | // L(k)
|
|---|
| 557 | //
|
|---|
| 558 | kp1 = k+1;
|
|---|
| 559 | km1 = k-1;
|
|---|
| 560 | v = work0[k];
|
|---|
| 561 | for(i_=kp1; i_<=n;i_++)
|
|---|
| 562 | {
|
|---|
| 563 | work0[i_] = work0[i_] + v*l[i_,k];
|
|---|
| 564 | }
|
|---|
| 565 | v = work0[km1];
|
|---|
| 566 | for(i_=kp1; i_<=n;i_++)
|
|---|
| 567 | {
|
|---|
| 568 | work0[i_] = work0[i_] + v*l[i_,km1];
|
|---|
| 569 | }
|
|---|
| 570 |
|
|---|
| 571 | //
|
|---|
| 572 | // P(k)
|
|---|
| 573 | //
|
|---|
| 574 | v = work0[k];
|
|---|
| 575 | work0[k] = work0[-pivots[k]];
|
|---|
| 576 | work0[-pivots[k]] = v;
|
|---|
| 577 |
|
|---|
| 578 | //
|
|---|
| 579 | // Next k
|
|---|
| 580 | //
|
|---|
| 581 | k = k-2;
|
|---|
| 582 | }
|
|---|
| 583 | }
|
|---|
| 584 | }
|
|---|
| 585 | }
|
|---|
| 586 | }
|
|---|
| 587 |
|
|---|
| 588 | //
|
|---|
| 589 | // Quick return if possible
|
|---|
| 590 | //
|
|---|
| 591 | rcond = 0;
|
|---|
| 592 | if( n==0 )
|
|---|
| 593 | {
|
|---|
| 594 | rcond = 1;
|
|---|
| 595 | return;
|
|---|
| 596 | }
|
|---|
| 597 | if( (double)(anorm)==(double)(0) )
|
|---|
| 598 | {
|
|---|
| 599 | return;
|
|---|
| 600 | }
|
|---|
| 601 |
|
|---|
| 602 | //
|
|---|
| 603 | // Estimate the 1-norm of inv(A).
|
|---|
| 604 | //
|
|---|
| 605 | kase = 0;
|
|---|
| 606 | while( true )
|
|---|
| 607 | {
|
|---|
| 608 | estnorm.iterativeestimate1norm(n, ref work1, ref work0, ref iwork, ref ainvnm, ref kase);
|
|---|
| 609 | if( kase==0 )
|
|---|
| 610 | {
|
|---|
| 611 | break;
|
|---|
| 612 | }
|
|---|
| 613 | ssolve.solvesystemldlt(ref l, ref pivots, work0, n, isupper, ref work2);
|
|---|
| 614 | for(i_=1; i_<=n;i_++)
|
|---|
| 615 | {
|
|---|
| 616 | work0[i_] = work2[i_];
|
|---|
| 617 | }
|
|---|
| 618 | }
|
|---|
| 619 |
|
|---|
| 620 | //
|
|---|
| 621 | // Compute the estimate of the reciprocal condition number.
|
|---|
| 622 | //
|
|---|
| 623 | if( (double)(ainvnm)!=(double)(0) )
|
|---|
| 624 | {
|
|---|
| 625 | v = 1/ainvnm;
|
|---|
| 626 | rcond = v/anorm;
|
|---|
| 627 | }
|
|---|
| 628 | }
|
|---|
| 629 | }
|
|---|
| 630 | }
|
|---|
