1 package Jama;
2 import Jama.util.*;
3
4 /** Singular Value Decomposition.
5 <P>
6 For an m-by-n matrix A with m >= n, the singular value decomposition is
7 an m-by-n orthogonal matrix U, an n-by-n diagonal matrix S, and
8 an n-by-n orthogonal matrix V so that A = U*S*V'.
9 <P>
10 The singular values, sigma[k] = S[k][k], are ordered so that
11 sigma[0] >= sigma[1] >= ... >= sigma[n-1].
12 <P>
13 The singular value decompostion always exists, so the constructor will
14 never fail. The matrix condition number and the effective numerical
15 rank can be computed from this decomposition.
16 @version $Id: SingularValueDecomposition.java,v 1.1.1.1 2010/11/30 21:32:00 jeremy Exp $
17 */
18
19 public class SingularValueDecomposition implements java.io.Serializable {
20
21 /* ------------------------
22 Class variables
23 * ------------------------ */
24
25 /** Arrays for internal storage of U and V.
26 @serial internal storage of U.
27 @serial internal storage of V.
28 */
29 private double[][] U, V;
30
31 /** Array for internal storage of singular values.
32 @serial internal storage of singular values.
33 */
34 private double[] s;
35
36 /** Row and column dimensions.
37 @serial row dimension.
38 @serial column dimension.
39 */
40 private int m, n;
41
42 /* ------------------------
43 Constructor
44 * ------------------------ */
45
46 /** Construct the singular value decomposition
47 @param Arg Rectangular matrix
48 */
49
50 public SingularValueDecomposition (Matrix Arg) {
51
52 // Derived from LINPACK code.
53 // Initialize.
54 double[][] A = Arg.getArrayCopy();
55 m = Arg.getRowDimension();
56 n = Arg.getColumnDimension();
57 int nu = Math.min(m,n);
58 s = new double [Math.min(m+1,n)];
59 U = new double [m][nu];
60 V = new double [n][n];
61 double[] e = new double [n];
62 double[] work = new double [m];
63 boolean wantu = true;
64 boolean wantv = true;
65
66 // Reduce A to bidiagonal form, storing the diagonal elements
67 // in s and the super-diagonal elements in e.
68
69 int nct = Math.min(m-1,n);
70 int nrt = Math.max(0,Math.min(n-2,m));
71 for (int k = 0; k < Math.max(nct,nrt); k++) {
72 if (k < nct) {
73
74 // Compute the transformation for the k-th column and
75 // place the k-th diagonal in s[k].
76 // Compute 2-norm of k-th column without under/overflow.
77 s[k] = 0;
78 for (int i = k; i < m; i++) {
79 s[k] = Maths.hypot(s[k],A[i][k]);
80 }
81 if (s[k] != 0.0) {
82 if (A[k][k] < 0.0) {
83 s[k] = -s[k];
84 }
85 for (int i = k; i < m; i++) {
86 A[i][k] /= s[k];
87 }
88 A[k][k] += 1.0;
89 }
90 s[k] = -s[k];
91 }
92 for (int j = k+1; j < n; j++) {
93 if ((k < nct) & (s[k] != 0.0)) {
94
95 // Apply the transformation.
96
97 double t = 0;
98 for (int i = k; i < m; i++) {
99 t += A[i][k]*A[i][j];
100 }
101 t = -t/A[k][k];
102 for (int i = k; i < m; i++) {
103 A[i][j] += t*A[i][k];
104 }
105 }
106
107 // Place the k-th row of A into e for the
108 // subsequent calculation of the row transformation.
109
110 e[j] = A[k][j];
111 }
112 if (wantu & (k < nct)) {
113
114 // Place the transformation in U for subsequent back
115 // multiplication.
116
117 for (int i = k; i < m; i++) {
118 U[i][k] = A[i][k];
119 }
120 }
121 if (k < nrt) {
122
123 // Compute the k-th row transformation and place the
124 // k-th super-diagonal in e[k].
125 // Compute 2-norm without under/overflow.
126 e[k] = 0;
127 for (int i = k+1; i < n; i++) {
128 e[k] = Maths.hypot(e[k],e[i]);
129 }
130 if (e[k] != 0.0) {
131 if (e[k+1] < 0.0) {
132 e[k] = -e[k];
133 }
134 for (int i = k+1; i < n; i++) {
135 e[i] /= e[k];
136 }
137 e[k+1] += 1.0;
138 }
139 e[k] = -e[k];
140 if ((k+1 < m) & (e[k] != 0.0)) {
141
142 // Apply the transformation.
143
144 for (int i = k+1; i < m; i++) {
145 work[i] = 0.0;
146 }
147 for (int j = k+1; j < n; j++) {
148 for (int i = k+1; i < m; i++) {
149 work[i] += e[j]*A[i][j];
150 }
151 }
152 for (int j = k+1; j < n; j++) {
153 double t = -e[j]/e[k+1];
154 for (int i = k+1; i < m; i++) {
155 A[i][j] += t*work[i];
156 }
157 }
158 }
159 if (wantv) {
160
161 // Place the transformation in V for subsequent
162 // back multiplication.
163
164 for (int i = k+1; i < n; i++) {
165 V[i][k] = e[i];
166 }
167 }
168 }
169 }
170
171 // Set up the final bidiagonal matrix or order p.
172
173 int p = Math.min(n,m+1);
174 if (nct < n) {
175 s[nct] = A[nct][nct];
176 }
177 if (m < p) {
178 s[p-1] = 0.0;
179 }
180 if (nrt+1 < p) {
181 e[nrt] = A[nrt][p-1];
182 }
183 e[p-1] = 0.0;
184
185 // If required, generate U.
186
187 if (wantu) {
188 for (int j = nct; j < nu; j++) {
189 for (int i = 0; i < m; i++) {
190 U[i][j] = 0.0;
191 }
192 U[j][j] = 1.0;
193 }
194 for (int k = nct-1; k >= 0; k--) {
195 if (s[k] != 0.0) {
196 for (int j = k+1; j < nu; j++) {
197 double t = 0;
198 for (int i = k; i < m; i++) {
199 t += U[i][k]*U[i][j];
200 }
201 t = -t/U[k][k];
202 for (int i = k; i < m; i++) {
203 U[i][j] += t*U[i][k];
204 }
205 }
206 for (int i = k; i < m; i++ ) {
207 U[i][k] = -U[i][k];
208 }
209 U[k][k] = 1.0 + U[k][k];
210 for (int i = 0; i < k-1; i++) {
211 U[i][k] = 0.0;
212 }
213 } else {
214 for (int i = 0; i < m; i++) {
215 U[i][k] = 0.0;
216 }
217 U[k][k] = 1.0;
218 }
219 }
220 }
221
222 // If required, generate V.
223
224 if (wantv) {
225 for (int k = n-1; k >= 0; k--) {
226 if ((k < nrt) & (e[k] != 0.0)) {
227 for (int j = k+1; j < nu; j++) {
228 double t = 0;
229 for (int i = k+1; i < n; i++) {
230 t += V[i][k]*V[i][j];
231 }
232 t = -t/V[k+1][k];
233 for (int i = k+1; i < n; i++) {
234 V[i][j] += t*V[i][k];
235 }
236 }
237 }
238 for (int i = 0; i < n; i++) {
239 V[i][k] = 0.0;
240 }
241 V[k][k] = 1.0;
242 }
243 }
244
245 // Main iteration loop for the singular values.
246
247 int pp = p-1;
248 int iter = 0;
249 double eps = Math.pow(2.0,-52.0);
250 while (p > 0) {
251 int k,kase;
252
253 // Here is where a test for too many iterations would go.
254
255 // This section of the program inspects for
256 // negligible elements in the s and e arrays. On
257 // completion the variables kase and k are set as follows.
258
259 // kase = 1 if s(p) and e[k-1] are negligible and k<p
260 // kase = 2 if s(k) is negligible and k<p
261 // kase = 3 if e[k-1] is negligible, k<p, and
262 // s(k), ..., s(p) are not negligible (qr step).
263 // kase = 4 if e(p-1) is negligible (convergence).
264
265 for (k = p-2; k >= -1; k--) {
266 if (k == -1) {
267 break;
268 }
269 if (Math.abs(e[k]) <= eps*(Math.abs(s[k]) + Math.abs(s[k+1]))) {
270 e[k] = 0.0;
271 break;
272 }
273 }
274 if (k == p-2) {
275 kase = 4;
276 } else {
277 int ks;
278 for (ks = p-1; ks >= k; ks--) {
279 if (ks == k) {
280 break;
281 }
282 double t = (ks != p ? Math.abs(e[ks]) : 0.) +
283 (ks != k+1 ? Math.abs(e[ks-1]) : 0.);
284 if (Math.abs(s[ks]) <= eps*t) {
285 s[ks] = 0.0;
286 break;
287 }
288 }
289 if (ks == k) {
290 kase = 3;
291 } else if (ks == p-1) {
292 kase = 1;
293 } else {
294 kase = 2;
295 k = ks;
296 }
297 }
298 k++;
299
300 // Perform the task indicated by kase.
301
302 switch (kase) {
303
304 // Deflate negligible s(p).
305
306 case 1: {
307 double f = e[p-2];
308 e[p-2] = 0.0;
309 for (int j = p-2; j >= k; j--) {
310 double t = Maths.hypot(s[j],f);
311 double cs = s[j]/t;
312 double sn = f/t;
313 s[j] = t;
314 if (j != k) {
315 f = -sn*e[j-1];
316 e[j-1] = cs*e[j-1];
317 }
318 if (wantv) {
319 for (int i = 0; i < n; i++) {
320 t = cs*V[i][j] + sn*V[i][p-1];
321 V[i][p-1] = -sn*V[i][j] + cs*V[i][p-1];
322 V[i][j] = t;
323 }
324 }
325 }
326 }
327 break;
328
329 // Split at negligible s(k).
330
331 case 2: {
332 double f = e[k-1];
333 e[k-1] = 0.0;
334 for (int j = k; j < p; j++) {
335 double t = Maths.hypot(s[j],f);
336 double cs = s[j]/t;
337 double sn = f/t;
338 s[j] = t;
339 f = -sn*e[j];
340 e[j] = cs*e[j];
341 if (wantu) {
342 for (int i = 0; i < m; i++) {
343 t = cs*U[i][j] + sn*U[i][k-1];
344 U[i][k-1] = -sn*U[i][j] + cs*U[i][k-1];
345 U[i][j] = t;
346 }
347 }
348 }
349 }
350 break;
351
352 // Perform one qr step.
353
354 case 3: {
355
356 // Calculate the shift.
357
358 double scale = Math.max(Math.max(Math.max(Math.max(
359 Math.abs(s[p-1]),Math.abs(s[p-2])),Math.abs(e[p-2])),
360 Math.abs(s[k])),Math.abs(e[k]));
361 double sp = s[p-1]/scale;
362 double spm1 = s[p-2]/scale;
363 double epm1 = e[p-2]/scale;
364 double sk = s[k]/scale;
365 double ek = e[k]/scale;
366 double b = ((spm1 + sp)*(spm1 - sp) + epm1*epm1)/2.0;
367 double c = (sp*epm1)*(sp*epm1);
368 double shift = 0.0;
369 if ((b != 0.0) | (c != 0.0)) {
370 shift = Math.sqrt(b*b + c);
371 if (b < 0.0) {
372 shift = -shift;
373 }
374 shift = c/(b + shift);
375 }
376 double f = (sk + sp)*(sk - sp) + shift;
377 double g = sk*ek;
378
379 // Chase zeros.
380
381 for (int j = k; j < p-1; j++) {
382 double t = Maths.hypot(f,g);
383 double cs = f/t;
384 double sn = g/t;
385 if (j != k) {
386 e[j-1] = t;
387 }
388 f = cs*s[j] + sn*e[j];
389 e[j] = cs*e[j] - sn*s[j];
390 g = sn*s[j+1];
391 s[j+1] = cs*s[j+1];
392 if (wantv) {
393 for (int i = 0; i < n; i++) {
394 t = cs*V[i][j] + sn*V[i][j+1];
395 V[i][j+1] = -sn*V[i][j] + cs*V[i][j+1];
396 V[i][j] = t;
397 }
398 }
399 t = Maths.hypot(f,g);
400 cs = f/t;
401 sn = g/t;
402 s[j] = t;
403 f = cs*e[j] + sn*s[j+1];
404 s[j+1] = -sn*e[j] + cs*s[j+1];
405 g = sn*e[j+1];
406 e[j+1] = cs*e[j+1];
407 if (wantu && (j < m-1)) {
408 for (int i = 0; i < m; i++) {
409 t = cs*U[i][j] + sn*U[i][j+1];
410 U[i][j+1] = -sn*U[i][j] + cs*U[i][j+1];
411 U[i][j] = t;
412 }
413 }
414 }
415 e[p-2] = f;
416 iter = iter + 1;
417 }
418 break;
419
420 // Convergence.
421
422 case 4: {
423
424 // Make the singular values positive.
425
426 if (s[k] <= 0.0) {
427 s[k] = (s[k] < 0.0 ? -s[k] : 0.0);
428 if (wantv) {
429 for (int i = 0; i <= pp; i++) {
430 V[i][k] = -V[i][k];
431 }
432 }
433 }
434
435 // Order the singular values.
436
437 while (k < pp) {
438 if (s[k] >= s[k+1]) {
439 break;
440 }
441 double t = s[k];
442 s[k] = s[k+1];
443 s[k+1] = t;
444 if (wantv && (k < n-1)) {
445 for (int i = 0; i < n; i++) {
446 t = V[i][k+1]; V[i][k+1] = V[i][k]; V[i][k] = t;
447 }
448 }
449 if (wantu && (k < m-1)) {
450 for (int i = 0; i < m; i++) {
451 t = U[i][k+1]; U[i][k+1] = U[i][k]; U[i][k] = t;
452 }
453 }
454 k++;
455 }
456 iter = 0;
457 p--;
458 }
459 break;
460 }
461 }
462 }
463
464 /* ------------------------
465 Public Methods
466 * ------------------------ */
467
468 /** Return the left singular vectors
469 @return U
470 */
471
472 public Matrix getU () {
473 return new Matrix(U,m,Math.min(m+1,n));
474 }
475
476 /** Return the right singular vectors
477 @return V
478 */
479
480 public Matrix getV () {
481 return new Matrix(V,n,n);
482 }
483
484 /** Return the one-dimensional array of singular values
485 @return diagonal of S.
486 */
487
488 public double[] getSingularValues () {
489 return s;
490 }
491
492 /** Return the diagonal matrix of singular values
493 @return S
494 */
495
496 public Matrix getS () {
497 Matrix X = new Matrix(n,n);
498 double[][] S = X.getArray();
499 for (int i = 0; i < n; i++) {
500 for (int j = 0; j < n; j++) {
501 S[i][j] = 0.0;
502 }
503 S[i][i] = this.s[i];
504 }
505 return X;
506 }
507
508 /** Two norm
509 @return max(S)
510 */
511
512 public double norm2 () {
513 return s[0];
514 }
515
516 /** Two norm condition number
517 @return max(S)/min(S)
518 */
519
520 public double cond () {
521 return s[0]/s[Math.min(m,n)-1];
522 }
523
524 /** Effective numerical matrix rank
525 @return Number of nonnegligible singular values.
526 */
527
528 public int rank () {
529 double eps = Math.pow(2.0,-52.0);
530 double tol = Math.max(m,n)*s[0]*eps;
531 int r = 0;
532 for (int i = 0; i < s.length; i++) {
533 if (s[i] > tol) {
534 r++;
535 }
536 }
537 return r;
538 }
539 }