content: package lib.comirva.audio;
  
  
  /**
   * 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 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;
              }
          }
      }
  
  /* ------------------------
     Public Methods
   * ------------------------ */
  
      /**
       * Return the left singular vectors
       *
       * @return U
       */
  
      public Matrix getU() {
          return new Matrix(U, m, Math.min(m + 1, n));
      }
  
      /**
       * Return the right singular vectors
       *
       * @return V
       */
  
      public Matrix getV() {
          return 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(n, n);
          double[][] S = X.getArray();
          for (int i = 0; i < n; i++) {
              for (int j = 0; j < n; j++) {
                  S[i][j] = 0.0;
              }
              S[i][i] = this.s[i];
          }
          return X;
      }
  
      /**
       * 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 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;
      }
  }
  