/**
* Effective numerical matrix rank
*
* @return Number of nonnegligible singular values.
*/
public int rank() {
double eps = Math.pow(2.0, -52.0);
double tol = Math.max(m, n) * s[0] * eps;
int r = 0;
for (int i = 0; i < s.length; i++) {
if (s[i] > tol) {
r++;
}
}
return r;
}
/**
* Construct the singular value decomposition Structure to access U, S and V.
*
* @param Arg Rectangular matrix
*/
public SingularValueDecomposition(Matrix Arg) {
// Derived from LINPACK code.
// Initialize.
double[][] A = Arg.getArrayCopy();
m = Arg.getRowDimension();
n = Arg.getColumnDimension();
/* Apparently the failing cases are only a proper subset of (m<n),
so let's not throw error. Correct fix to come later?
if (m<n) {
throw new IllegalArgumentException("Jama SVD only works for m >= n"); }
*/
int nu = Math.min(m, n);
s = new double[Math.min(m + 1, n)];
U = new double[m][nu];
V = new double[n][n];
double[] e = new double[n];
double[] work = new double[m];
boolean wantu = true;
boolean wantv = true;
// Reduce A to bidiagonal form, storing the diagonal elements
// in s and the super-diagonal elements in e.
int nct = Math.min(m - 1, n);
int nrt = Math.max(0, Math.min(n - 2, m));
for (int k = 0; k < Math.max(nct, nrt); k++) {
if (k < nct) {
// Compute the transformation for the k-th column and
// place the k-th diagonal in s[k].
// Compute 2-norm of k-th column without under/overflow.
s[k] = 0;
for (int i = k; i < m; i++) {
s[k] = Maths.hypot(s[k], A[i][k]);
}
if (s[k] != 0.0) {
if (A[k][k] < 0.0) {
s[k] = -s[k];
}
for (int i = k; i < m; i++) {
A[i][k] /= s[k];
}
A[k][k] += 1.0;
}
s[k] = -s[k];
}
for (int j = k + 1; j < n; j++) {
if ((k < nct) & (s[k] != 0.0)) {
// Apply the transformation.
double t = 0;
for (int i = k; i < m; i++) {
t += A[i][k] * A[i][j];
}
t = -t / A[k][k];
for (int i = k; i < m; i++) {
A[i][j] += t * A[i][k];
}
}
// Place the k-th row of A into e for the
// subsequent calculation of the row transformation.
e[j] = A[k][j];
}
if (wantu & (k < nct)) {
// Place the transformation in U for subsequent back
// multiplication.
for (int i = k; i < m; i++) {
U[i][k] = A[i][k];
}
}
if (k < nrt) {
// Compute the k-th row transformation and place the
// k-th super-diagonal in e[k].
// Compute 2-norm without under/overflow.
e[k] = 0;
for (int i = k + 1; i < n; i++) {
e[k] = Maths.hypot(e[k], e[i]);
}
if (e[k] != 0.0) {
if (e[k + 1] < 0.0) {
e[k] = -e[k];
}
for (int i = k + 1; i < n; i++) {
e[i] /= e[k];
}
e[k + 1] += 1.0;
}
e[k] = -e[k];
if ((k + 1 < m) & (e[k] != 0.0)) {
// Apply the transformation.
for (int i = k + 1; i < m; i++) {
work[i] = 0.0;
}
for (int j = k + 1; j < n; j++) {
for (int i = k + 1; i < m; i++) {
work[i] += e[j] * A[i][j];
}
}
for (int j = k + 1; j < n; j++) {
double t = -e[j] / e[k + 1];
for (int i = k + 1; i < m; i++) {
A[i][j] += t * work[i];
}
}
}
if (wantv) {
// Place the transformation in V for subsequent
// back multiplication.
for (int i = k + 1; i < n; i++) {
V[i][k] = e[i];
}
}
}
}
// Set up the final bidiagonal matrix or order p.
int p = Math.min(n, m + 1);
if (nct < n) {
s[nct] = A[nct][nct];
}
if (m < p) {
s[p - 1] = 0.0;
}
if (nrt + 1 < p) {
e[nrt] = A[nrt][p - 1];
}
e[p - 1] = 0.0;
// If required, generate U.
if (wantu) {
for (int j = nct; j < nu; j++) {
for (int i = 0; i < m; i++) {
U[i][j] = 0.0;
}
U[j][j] = 1.0;
}
for (int k = nct - 1; k >= 0; k--) {
if (s[k] != 0.0) {
for (int j = k + 1; j < nu; j++) {
double t = 0;
for (int i = k; i < m; i++) {
t += U[i][k] * U[i][j];
}
t = -t / U[k][k];
for (int i = k; i < m; i++) {
U[i][j] += t * U[i][k];
}
}
for (int i = k; i < m; i++) {
U[i][k] = -U[i][k];
}
U[k][k] = 1.0 + U[k][k];
for (int i = 0; i < k - 1; i++) {
U[i][k] = 0.0;
}
} else {
for (int i = 0; i < m; i++) {
U[i][k] = 0.0;
}
U[k][k] = 1.0;
}
}
}
// If required, generate V.
if (wantv) {
for (int k = n - 1; k >= 0; k--) {
if ((k < nrt) & (e[k] != 0.0)) {
for (int j = k + 1; j < nu; j++) {
double t = 0;
for (int i = k + 1; i < n; i++) {
t += V[i][k] * V[i][j];
}
t = -t / V[k + 1][k];
for (int i = k + 1; i < n; i++) {
V[i][j] += t * V[i][k];
}
}
}
for (int i = 0; i < n; i++) {
V[i][k] = 0.0;
}
V[k][k] = 1.0;
}
}
// Main iteration loop for the singular values.
int pp = p - 1;
int iter = 0;
double eps = Math.pow(2.0, -52.0);
double tiny = Math.pow(2.0, -966.0);
while (p > 0) {
int k, kase;
// Here is where a test for too many iterations would go.
// This section of the program inspects for
// negligible elements in the s and e arrays. On
// completion the variables kase and k are set as follows.
// kase = 1 if s(p) and e[k-1] are negligible and k<p
// kase = 2 if s(k) is negligible and k<p
// kase = 3 if e[k-1] is negligible, k<p, and
// s(k), ..., s(p) are not negligible (qr step).
// kase = 4 if e(p-1) is negligible (convergence).
for (k = p - 2; k >= -1; k--) {
if (k == -1) {
break;
}
if (Math.abs(e[k]) <= tiny + eps * (Math.abs(s[k]) + Math.abs(s[k + 1]))) {
e[k] = 0.0;
break;
}
}
if (k == p - 2) {
kase = 4;
} else {
int ks;
for (ks = p - 1; ks >= k; ks--) {
if (ks == k) {
break;
}
double t = (ks != p ? Math.abs(e[ks]) : 0.) + (ks != k + 1 ? Math.abs(e[ks - 1]) : 0.);
if (Math.abs(s[ks]) <= tiny + eps * t) {
s[ks] = 0.0;
break;
}
}
if (ks == k) {
kase = 3;
} else if (ks == p - 1) {
kase = 1;
} else {
kase = 2;
k = ks;
}
}
k++;
// Perform the task indicated by kase.
switch (kase) {
// Deflate negligible s(p).
case 1:
{
double f = e[p - 2];
e[p - 2] = 0.0;
for (int j = p - 2; j >= k; j--) {
double t = Maths.hypot(s[j], f);
double cs = s[j] / t;
double sn = f / t;
s[j] = t;
if (j != k) {
f = -sn * e[j - 1];
e[j - 1] = cs * e[j - 1];
}
if (wantv) {
for (int i = 0; i < n; i++) {
t = cs * V[i][j] + sn * V[i][p - 1];
V[i][p - 1] = -sn * V[i][j] + cs * V[i][p - 1];
V[i][j] = t;
}
}
}
}
break;
// Split at negligible s(k).
case 2:
{
double f = e[k - 1];
e[k - 1] = 0.0;
for (int j = k; j < p; j++) {
double t = Maths.hypot(s[j], f);
double cs = s[j] / t;
double sn = f / t;
s[j] = t;
f = -sn * e[j];
e[j] = cs * e[j];
if (wantu) {
for (int i = 0; i < m; i++) {
t = cs * U[i][j] + sn * U[i][k - 1];
U[i][k - 1] = -sn * U[i][j] + cs * U[i][k - 1];
U[i][j] = t;
}
}
}
}
break;
// Perform one qr step.
case 3:
{
// Calculate the shift.
double scale =
Math.max(
Math.max(
Math.max(
Math.max(Math.abs(s[p - 1]), Math.abs(s[p - 2])), Math.abs(e[p - 2])),
Math.abs(s[k])),
Math.abs(e[k]));
double sp = s[p - 1] / scale;
double spm1 = s[p - 2] / scale;
double epm1 = e[p - 2] / scale;
double sk = s[k] / scale;
double ek = e[k] / scale;
double b = ((spm1 + sp) * (spm1 - sp) + epm1 * epm1) / 2.0;
double c = (sp * epm1) * (sp * epm1);
double shift = 0.0;
if ((b != 0.0) | (c != 0.0)) {
shift = Math.sqrt(b * b + c);
if (b < 0.0) {
shift = -shift;
}
shift = c / (b + shift);
}
double f = (sk + sp) * (sk - sp) + shift;
double g = sk * ek;
// Chase zeros.
for (int j = k; j < p - 1; j++) {
double t = Maths.hypot(f, g);
double cs = f / t;
double sn = g / t;
if (j != k) {
e[j - 1] = t;
}
f = cs * s[j] + sn * e[j];
e[j] = cs * e[j] - sn * s[j];
g = sn * s[j + 1];
s[j + 1] = cs * s[j + 1];
if (wantv) {
for (int i = 0; i < n; i++) {
t = cs * V[i][j] + sn * V[i][j + 1];
V[i][j + 1] = -sn * V[i][j] + cs * V[i][j + 1];
V[i][j] = t;
}
}
t = Maths.hypot(f, g);
cs = f / t;
sn = g / t;
s[j] = t;
f = cs * e[j] + sn * s[j + 1];
s[j + 1] = -sn * e[j] + cs * s[j + 1];
g = sn * e[j + 1];
e[j + 1] = cs * e[j + 1];
if (wantu && (j < m - 1)) {
for (int i = 0; i < m; i++) {
t = cs * U[i][j] + sn * U[i][j + 1];
U[i][j + 1] = -sn * U[i][j] + cs * U[i][j + 1];
U[i][j] = t;
}
}
}
e[p - 2] = f;
iter = iter + 1;
}
break;
// Convergence.
case 4:
{
// Make the singular values positive.
if (s[k] <= 0.0) {
s[k] = (s[k] < 0.0 ? -s[k] : 0.0);
if (wantv) {
for (int i = 0; i <= pp; i++) {
V[i][k] = -V[i][k];
}
}
}
// Order the singular values.
while (k < pp) {
if (s[k] >= s[k + 1]) {
break;
}
double t = s[k];
s[k] = s[k + 1];
s[k + 1] = t;
if (wantv && (k < n - 1)) {
for (int i = 0; i < n; i++) {
t = V[i][k + 1];
V[i][k + 1] = V[i][k];
V[i][k] = t;
}
}
if (wantu && (k < m - 1)) {
for (int i = 0; i < m; i++) {
t = U[i][k + 1];
U[i][k + 1] = U[i][k];
U[i][k] = t;
}
}
k++;
}
iter = 0;
p--;
}
break;
}
}
}
/**
* Two norm condition number
*
* @return max(S)/min(S)
*/
public double cond() {
return s[0] / s[Math.min(m, n) - 1];
}
/**
* Return the left singular vectors
*
* @return U
*/
public Matrix getU() {
return new Matrix(U, m, Math.min(m + 1, n));
}