[Fwd: Re: [Fwd: SVD woe revisited (LINPACK derived possible remedy) final fully functional 100% backward compatible]]
- Subject: [Fwd: Re: [Fwd: SVD woe revisited (LINPACK derived possible remedy) final fully functional 100% backward compatible]]
- From: Ron Boisvert <boisvert@nist.gov>
- Date: Wed, 12 Dec 2007 16:05:01 -0500
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-------- Original Message --------
Subject: Re: [Fwd: SVD woe revisited (LINPACK derived possible remedy)
final fully functional 100% backward compatible]
Date: Wed, 12 Dec 2007 16:00:52 -0500
From: Sione <sionep@xtra.co.nz>
Reply-To: sionep@xtra.co.nz
To: boisvert@nist.gov
References: <475FFB65.9060900@nist.gov>
Andreas,
Thanks for sharing your code. As a matlab user myself, I find your SVD
modification useful, since matlab SVD does both m <= n and m > n.
Cheers,
Sione.
Ron Boisvert wrote:
>
>
> -------- Original Message --------
> Subject: SVD woe revisited (LINPACK derived possible remedy) final
> fully functional 100% backward compatible
> Date: Wed, 12 Dec 2007 10:11:07 -0500
> From: Andreas Kyrmegalos <andreask1@vivodinet.gr>
> Reply-To: andreask1@vivodinet.gr
> To: boisvert@nist.gov
>
>
>
> Hello again,
> I have updated the code. The old constructor has returned, and the
> get commands are compatible with the old output.
> I have verified results for matrices sized 4:1,4:2,4:3,3:3,3:4,2:5,1:3
> As a friendly reminder do note that:
>
> a) if m<n I=A*A+
> b) if m=n I=A*A+=A+*A
> c) if m>n I=A+*A
>
> I believe that at this point the code returns proper results for all
> cases.
>
> Best regards,
> Andreas
>
>
> ------------------------------------------------------------------------
>
> package Jama;
>
> import Jama.util.*;
>
> /** Singular Value Decomposition.
> <P>
> For an m-by-n matrix A, the singular value decomposition is
> an m-by-(m or n) orthogonal matrix U, a (m or n)-by-n diagonal matrix S, and
> an n-by-n orthogonal matrix V so that A = U*S*V'.
> <P>
> The singular values, sigma[k] = S[k][k], are ordered so that
> sigma[0] >= sigma[1] >= ... >= sigma[n-1].
> <P>
> The singular value decompostion always exists, so the constructor will
> never fail. The matrix condition number and the effective numerical
> rank can be computed from this decomposition.
> */
>
> public class SingularValueDecomposition implements java.io.Serializable {
>
> /* ------------------------
> Class variables
> * ------------------------ */
>
> /** Arrays for internal storage of U and V.
> @serial internal storage of U.
> @serial internal storage of V.
> */
> private double[][] U, V;
>
> /** Array for internal storage of singular values.
> @serial internal storage of singular values.
> */
> private double[] s;
>
> /** Row and column dimensions.
> @serial row dimension.
> @serial column dimension.
> @serial U column dimension.
> */
> private int m, n, ncu;
>
> /** Column specification of matrix U
> @serial U column dimension toggle
> */
>
> private boolean thin;
>
> /* ------------------------
> Old Constructor
> * ------------------------ */
> /** Construct the singular value decomposition
> @param Arg Rectangular matrix
> @return Structure to access U, S and V.
> */
>
> public SingularValueDecomposition (Matrix Arg) {
> this(Arg,true,true,true);
> }
>
> /* ------------------------
> Constructor
> * ------------------------ */
>
> /** Construct the singular value decomposition
> @param Arg Rectangular matrix
> @param thin If true U is economy sized
> @param wantu If true generate the U matrix
> @param wantv If true generate the V matrix
> @return Structure to access U, S and V.
> */
>
> public SingularValueDecomposition (Matrix Arg, boolean thin, boolean wantu,
> boolean wantv) {
>
> // Derived from LINPACK code.
> // Initialize.
> double[][] A = Arg.getArrayCopy();
> m = Arg.getRowDimension();
> n = Arg.getColumnDimension();
> this.thin = thin;
>
> ncu = thin?Math.min(m,n):m;
> s = new double [Math.min(m+1,n)];
> if (wantu) U = new double [m][ncu];
> if (wantv) V = new double [n][n];
> double[] e = new double [n];
> double[] work = new double [m];
>
> // 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));
> int lu = Math.max(nct,nrt);
> for (int k = 0; k < lu; 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 < ncu; 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 < ncu; 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;
> 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 < n; 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++;
> }
> 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 < n; 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;
> }
> }
> A = null;
> }
>
> /* ------------------------
> Public Methods
> * ------------------------ */
>
> /** Return the left singular vectors
> @return U
> */
>
> public Matrix getU () {
> return U==null?null:(new Matrix(U,m,m>=n?(thin?Math.min(m+1,n):ncu):ncu));
> }
>
> /** Return the right singular vectors
> @return V
> */
>
> public Matrix getV () {
> return V==null?null:new Matrix(V,n,n);
> }
>
> /** Return the one-dimensional array of singular values
> @return diagonal of S.
> */
>
> public double[] getSingularValues () {
> return s;
> }
>
> /** Return the diagonal matrix of singular values
> @return S
> */
>
> public Matrix getS () {
> Matrix X = new Matrix(m>=n?(thin?n:ncu):ncu,n);
> double[][] S = X.getArray();
> for (int i = Math.min(m,n)-1; i>=0; i--)
> S[i][i] = s[i];
> return X;
> }
>
> /** Return the diagonal matrix of the reciprocals of the singular values
> @return S+
> */
>
> public Matrix getreciprocalS () {
> Matrix X = new Matrix(n,m>=n?(thin?n:ncu):ncu);
> double[][] S = X.getArray();
> for (int i = Math.min(m,n)-1; i>=0; i--)
> S[i][i] = s[i]==0.0?0.0:1.0/s[i];
> return X;
> }
>
> /** Return the Moore-Penrose (generalized) inverse
> * Slightly modified version of Kim van der Linde's code
> @param omit if true tolerance based omitting of negligible singular values
> @return A+
> */
>
> public Matrix inverse(boolean omit) {
> double[][] inverse = new double[n][m];
> if(rank()> 0) {
> double[] reciprocalS = new double[s.length];
> if (omit) {
> double tol = Math.max(m,n)*s[0]*Math.pow(2.0,-52.0);
> for (int i = s.length-1;i>=0;i--)
> reciprocalS[i] = Math.abs(s[i])<tol?0.0:1.0/s[i];
> }
> else
> for (int i=s.length-1;i>=0;i--)
> reciprocalS[i] = s[i]==0.0?0.0:1.0/s[i];
> int min = Math.min(n, ncu);
> for (int i = n-1; i >= 0; i--)
> for (int j = m-1; j >= 0; j--)
> for (int k = min-1; k >= 0; k--)
> inverse[i][j] += V[i][k] * reciprocalS[k] * U[j][k];
> }
> return new Matrix(inverse);
> }
>
> /** Two norm
> @return max(S)
> */
>
> public double norm2 () {
> return s[0];
> }
>
> /** Two norm condition number
> @return max(S)/min(S)
> */
>
> public double cond () {
> return s[0]/s[Math.min(m,n)-1];
> }
>
> /** Effective numerical matrix rank
> @return Number of nonnegligible singular values.
> */
>
> public int rank () {
> double tol = Math.max(m,n)*s[0]*Math.pow(2.0,-52.0);
> int r = 0;
> for (int i = 0; i < s.length; i++) {
> if (s[i] > tol) {
> r++;
> }
> }
> return r;
> }
> }
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