package Jama;
import Jama.util.*;

   /** Singular Value Decomposition.
   <P>
   For an m-by-n matrix A with m >= n, the singular value decomposition is
   an m-by-n orthogonal matrix U, an 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.
   */
   private int m, n;

/* ------------------------
   Constructor
 * ------------------------ */

   /** Construct the singular value decomposition
   @param A    Rectangular matrix
   @return     Structure to access U, S and V.
   */

   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;